Problem editorial

Find in Mountain Array

HardBinary Search PatternsArrayBinary SearchInteractive

Editorial

Core insight

Find in Mountain Array · Binary Search Patterns

Core Insight for Find in Mountain Array

The core difficulty in LeetCode 1095 is that the entire array is not sorted in a single direction. Standard binary search requires monotonic data (always increasing or always decreasing). A mountain array is bitonic: it increases, reaches a peak, and then decreases.

However, the peak element acts as a boundary.

To the left of the peak (inclusive of the peak), the array is strictly ascending.
To the right of the peak, the array is strictly descending.

If we can identify the index of the peak, we can treat the problem as two independent binary search problems on sorted arrays.

To find the peak efficiently, we utilize the derivative (slope) logic used in finding extrema. By comparing mid and mid + 1, we can determine if we are on the ascending slope or the descending slope:

If arr[mid] < arr[mid + 1], we are climbing up; the peak is to the right.
If arr[mid] > arr[mid + 1], we are climbing down (or at the peak); the actual peak is at mid or to the left.

Visual Description: Imagine plotting the array values on a graph where the Y-axis is the value and X-axis is the index. The shape forms an inverted "V".

Phase 1 (Peak Finding): We probe the middle of the range. If the slope at that point is positive, we discard the left half. If the slope is negative, we discard the right half. This narrows the search space to the single highest point (the peak).
Phase 2 (Target Search): Once the peak index is fixed, the graph is split into a "Left Wing" (strictly increasing) and a "Right Wing" (strictly decreasing). We perform binary search on the Left Wing first. If the target is found, we return that index (since we need the minimum index). If not, we perform binary search on the Right Wing.

Deep dive

Full breakdown

Full reasoning, pitfalls, and implementation flow

LeetCode 1095: Find in Mountain Array Solution & Explanation

Problem Overview

TL;DR: We solve this by first locating the peak element's index using binary search, then performing two separate binary searches on the ascending and descending slopes to find the target.

The Find in Mountain Array problem asks us to search for a specific target value within a "mountain array." A mountain array is defined as an array that strictly increases to a peak element and then strictly decreases. We are not given direct access to the array; instead, we must interact with it via a MountainArray interface that allows us to retrieve the length and the value at a specific index.

The critical constraint in LeetCode 1095 is the limit on API calls: we are allowed a maximum of 100 calls to get(k). Since the array length can be up to 10,000, a linear scan is impossible, necessitating a logarithmic time complexity solution.

Brute Force Approach for Find in Mountain Array

The most straightforward approach to finding an element in an array is a linear scan. We would simply iterate through every index from 0 to n-1, call get(i), and check if the value equals the target.

Naive Pseudo-code:

text

function findInMountainArray(target, mountainArr):
    length = mountainArr.length()
    for i from 0 to length - 1:
        val = mountainArr.get(i)
        if val == target:
            return i
    return -1

Complexity Analysis:

Time Complexity: $O(N)$ , where $N$ is the length of the mountain array.
Space Complexity: $O(1)$ .

Why it fails: The problem strictly enforces an interaction limit. With $N$ up to $10,000$ , a linear scan requires up to 10,000 calls to MountainArray.get(). The limit is 100 calls. Since $100 \ll 10,000$ , this approach will immediately trigger a "Wrong Answer" or disqualification due to excessive API usage. We must minimize calls using an $O(\log N)$ strategy.

Algorithm Strategy: Binary Search Patterns

Our optimal strategy involves three distinct binary search operations performed in sequence. This ensures we stay well within the 100-call limit, as $3 \times \log_2(10000) \approx 3 \times 14 = 42$ calls max.

Find the Peak Index:
- Use the logic from the "Find Min/Max" subpattern.
- Maintain low and high pointers.
- Compare get(mid) with get(mid + 1).
- If get(mid) < get(mid + 1), the peak is strictly to the right (low = mid + 1).
- Otherwise, the peak is at mid or to the left (high = mid).
- The loop terminates when low == high, giving us the peakIndex.
Search the Ascending Part:
- Perform a standard "Order-Agnostic" binary search on the range [0, peakIndex].
- Since this segment is sorted in ascending order, if get(mid) < target, we move right. If get(mid) > target, we move left.
- If found, return the index immediately (this guarantees the minimum index).
Search the Descending Part:
- If the target wasn't found in the ascending part, perform binary search on the range [peakIndex + 1, length - 1].
- Note that this segment is sorted in descending order.
- Logic is inverted: if get(mid) < target, we move left (towards larger values). If get(mid) > target, we move right.
- If found, return the index. Otherwise, return -1.

Execution Flow

Let's trace the algorithm with mountainArr = [1, 2, 3, 4, 5, 3, 1] and target = 3.

Step 1: Find Peak Index

Range [0, 6]. mid = 3.
Compare arr[3] (4) and arr[4] (5).
Since $4 < 5$ , we are ascending. Peak is to the right. New range [4, 6].
mid = 5. Compare arr[5] (3) and arr[6] (1).
Since $3 > 1$ , we are descending. Peak is at or left of 5. New range [4, 5].
mid = 4. Compare arr[4] (5) and arr[5] (3).
Since $5 > 3$ , we are descending. Peak is at or left of 4. New range [4, 4].
low == high. Peak Index is 4 (Value is 5).

Step 2: Search Left (Ascending) [0, 4]

Range [0, 4]. Target is 3.
mid = 2. arr[2] is 3.
Match found!
Since we want the minimum index and we search the left side first, we return 2.

Note: If the target were 3 in the right half only (e.g., if the array were [1, 5, 3]), Step 2 would fail, and we would proceed to Step 3.

Proof of Correctness

The correctness relies on the Bitonic Property of the Mountain Array.

Peak Uniqueness: A mountain array has exactly one peak. The condition arr[i] < arr[i+1] is true for all $i < \text{peak}$ and false for all $i \geq \text{peak}$ . This monotonicity in the slope allows binary search to converge on the peak index correctly.
Coverage: By splitting the array at the peak, we cover the entire search space. The left side [0, peak] is strictly increasing, and the right side [peak+1, end] is strictly decreasing.
Min Index Requirement: Searching the left subarray first ensures that if the target appears twice (once on each slope), we return the smaller index.

Pattern Reuse Notes

The logic used in Find in Mountain Array is a specific application of binary search on modified sorted arrays. This pattern recurs in several famous interview questions:

LeetCode 33: Search in Rotated Sorted Array: Similar to the mountain array, a rotated array consists of two sorted segments. You must determine which segment is sorted to decide pointer movement.
LeetCode 81: Search in Rotated Sorted Array II: Handles duplicates in the rotated array context, requiring worst-case $O(N)$ fallback but utilizing similar pivot logic.
LeetCode 153: Find Minimum in Rotated Sorted Array: This is the direct inverse of finding the peak. You use binary search to find the point where the order breaks (the minimum).
LeetCode 162: Find Peak Element: This is exactly the first step of the Mountain Array solution. It asks you to find any peak in an unsorted array where adjacent elements are never equal.

Feynman coach · Before you peek at the code

Can you explain the core insight in your own words?

If you can explain it before reading code, you’re far more likely to recall it under interview pressure.

Loading your attempt…

Reference Implementation

C++ Solution for LeetCode 1095

cpp

1/**
2 * // This is the MountainArray's API interface.
3 * // You should not implement it, or speculate about its implementation
4 * class MountainArray {
5 *   public:
6 *     int get(int index);
7 *     int length();
8 * };
9 */
10
11class Solution {
12public:
13    int findInMountainArray(int target, MountainArray &mountainArr) {
14        int n = mountainArr.length();
15        
16        // 1. Find the peak index
17        int low = 0;
18        int high = n - 1;
19        while (low < high) {
20            int mid = low + (high - low) / 2;
21            if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
22                // Ascending part, peak is to the right
23                low = mid + 1;
24            } else {
25                // Descending part or peak, peak is left or here
26                high = mid;
27            }
28        }
29        int peakIndex = low;
30        
31        // 2. Search in the ascending part (left of peak)
32        low = 0;
33        high = peakIndex;
34        while (low <= high) {
35            int mid = low + (high - low) / 2;
36            int val = mountainArr.get(mid);
37            if (val == target) return mid;
38            if (val < target) {
39                low = mid + 1;
40            } else {
41                high = mid - 1;
42            }
43        }
44        
45        // 3. Search in the descending part (right of peak)
46        low = peakIndex + 1;
47        high = n - 1;
48        while (low <= high) {
49            int mid = low + (high - low) / 2;
50            int val = mountainArr.get(mid);
51            if (val == target) return mid;
52            if (val < target) {
53                // In descending order, smaller value means we are too far right
54                high = mid - 1;
55            } else {
56                // Larger value means we are too far left
57                low = mid + 1;
58            }
59        }
60        
61        return -1;
62    }
63};

Java Solution for LeetCode 1095

java

1/**
2 * // This is MountainArray's API interface.
3 * // You should not implement it, or speculate about its implementation
4 * interface MountainArray {
5 *     public int get(int index);
6 *     public int length();
7 * }
8 */
9 
10class Solution {
11    public int findInMountainArray(int target, MountainArray mountainArr) {
12        int length = mountainArr.length();
13        
14        // 1. Find the peak index
15        int low = 0;
16        int high = length - 1;
17        while (low < high) {
18            int mid = low + (high - low) / 2;
19            if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
20                // We are in the increasing section
21                low = mid + 1;
22            } else {
23                // We are in the decreasing section or at the peak
24                high = mid;
25            }
26        }
27        int peakIndex = low;
28        
29        // 2. Binary Search on the increasing subarray [0, peakIndex]
30        int result = binarySearch(mountainArr, target, 0, peakIndex, true);
31        if (result != -1) {
32            return result;
33        }
34        
35        // 3. Binary Search on the decreasing subarray [peakIndex + 1, length - 1]
36        return binarySearch(mountainArr, target, peakIndex + 1, length - 1, false);
37    }
38    
39    private int binarySearch(MountainArray arr, int target, int low, int high, boolean ascending) {
40        while (low <= high) {
41            int mid = low + (high - low) / 2;
42            int val = arr.get(mid);
43            
44            if (val == target) {
45                return mid;
46            }
47            
48            if (ascending) {
49                if (val < target) {
50                    low = mid + 1;
51                } else {
52                    high = mid - 1;
53                }
54            } else {
55                // Descending logic
56                if (val < target) {
57                    high = mid - 1; // Target is larger, move left
58                } else {
59                    low = mid + 1;  // Target is smaller, move right
60                }
61            }
62        }
63        return -1;
64    }
65}

Python Solution for LeetCode 1095

python

1# """
2# This is MountainArray's API interface.
3# You should not implement it, or speculate about its implementation
4# """
5#class MountainArray:
6#    def get(self, index: int) -> int:
7#    def length(self) -> int:
8
9class Solution:
10    def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
11        n = mountain_arr.length()
12        
13        # 1. Find the peak index
14        low, high = 0, n - 1
15        while low < high:
16            mid = (low + high) // 2
17            if mountain_arr.get(mid) < mountain_arr.get(mid + 1):
18                low = mid + 1
19            else:
20                high = mid
21        peak_index = low
22        
23        # 2. Search in the ascending part
24        low, high = 0, peak_index
25        while low <= high:
26            mid = (low + high) // 2
27            val = mountain_arr.get(mid)
28            if val == target:
29                return mid
30            elif val < target:
31                low = mid + 1
32            else:
33                high = mid - 1
34                
35        # 3. Search in the descending part
36        low, high = peak_index + 1, n - 1
37        while low <= high:
38            mid = (low + high) // 2
39            val = mountain_arr.get(mid)
40            if val == target:
41                return mid
42            elif val < target:
43                # In descending, smaller value means go left (to larger values)
44                high = mid - 1
45            else:
46                low = mid + 1
47                
48        return -1

AI Coach Insight

For Find in Mountain Array, focus on explaining why the structure works, not just naming Binary Search Patterns.

Complexity Analysis

Time Complexity: $O(\log N)$
- Finding the peak takes $O(\log N)$ .
- Searching the left part takes $O(\log N)$ .
- Searching the right part takes $O(\log N)$ .
- Total time is proportional to $3 \log N$ , which simplifies to $O(\log N)$ .
- With $N=10,000$ , this results in roughly 40-45 comparisons, well within the 100 call limit.
Space Complexity: $O(1)$
- We only use a few variables for pointers (low, high, mid, peak) and values. No recursion stack or auxiliary data structures are needed if implemented iteratively.

Related Patterns

Pattern: Binary Search Patterns
Subpattern: Binary Search - Find Min/Max in Rotated Sorted Array

This problem fits the Binary Search pattern because we are searching for a value in a structure that possesses sorted properties (monotonicity), albeit in two distinct segments. The subpattern "Find Min/Max" is applicable because the crucial first step is identifying the "peak" (maximum) of the mountain, which acts as the pivot point dividing the array into two searchable sorted halves.

Common Mistakes & FAQ

Why does the naive linear scan fail? The naive solution fails because the problem imposes a hard limit of 100 calls to get(). A linear scan requires $O(N)$ calls. Since $N$ can be up to 10,000, this drastically exceeds the limit, resulting in a Time Limit Exceeded or Wrong Answer verdict.

Mistake 1: Not handling the descending order correctly.

What: Applying standard binary search logic (val < target means low = mid + 1) to the right side of the mountain.
Why: The right side of the mountain is sorted in descending order (large to small).
Consequence: The pointers move in the wrong direction, and the target is never found even if it exists.

Mistake 2: Caching the peak value but not the index.

What: Finding the peak value correctly but failing to store its index to use as a boundary for subsequent searches.
Why: The algorithm requires the index to split the array into [0, peakIndex] and [peakIndex + 1, end].
Consequence: You cannot define the boundaries for the subsequent binary searches.

Mistake 3: Infinite loops during peak finding.

What: Using high = mid - 1 or low = mid incorrectly without adjusting the loop condition or mid calculation.
Why: Boundary conditions in binary search are subtle. If low and high are adjacent, mid might calculate to low, and if the logic doesn't advance pointers, the loop never terminates.
Consequence: Time Limit Exceeded (TLE).

Practise Find in Mountain Array and the problems around it

Find in Mountain Array is a Binary Search Patterns problem, in the Find Min/Max in Rotated Sorted Array subpattern. Once the approach above makes sense, write it yourself and then work through the neighbouring problems that reuse it.

Core insight

Find in Mountain Array · Binary Search Patterns

Core Insight for Find in Mountain Array

However, the peak element acts as a boundary.

To the left of the peak (inclusive of the peak), the array is strictly ascending.
To the right of the peak, the array is strictly descending.

If we can identify the index of the peak, we can treat the problem as two independent binary search problems on sorted arrays.

If arr[mid] < arr[mid + 1], we are climbing up; the peak is to the right.
If arr[mid] > arr[mid + 1], we are climbing down (or at the peak); the actual peak is at mid or to the left.

Visual Description: Imagine plotting the array values on a graph where the Y-axis is the value and X-axis is the index. The shape forms an inverted "V".

Phase 1 (Peak Finding): We probe the middle of the range. If the slope at that point is positive, we discard the left half. If the slope is negative, we discard the right half. This narrows the search space to the single highest point (the peak).
Phase 2 (Target Search): Once the peak index is fixed, the graph is split into a "Left Wing" (strictly increasing) and a "Right Wing" (strictly decreasing). We perform binary search on the Left Wing first. If the target is found, we return that index (since we need the minimum index). If not, we perform binary search on the Right Wing.

Deep dive

Full breakdown

Full reasoning, pitfalls, and implementation flow

LeetCode 1095: Find in Mountain Array Solution & Explanation

Problem Overview

TL;DR: We solve this by first locating the peak element's index using binary search, then performing two separate binary searches on the ascending and descending slopes to find the target.

Brute Force Approach for Find in Mountain Array

Naive Pseudo-code:

text

function findInMountainArray(target, mountainArr):
    length = mountainArr.length()
    for i from 0 to length - 1:
        val = mountainArr.get(i)
        if val == target:
            return i
    return -1

Complexity Analysis:

Time Complexity: $O(N)$ , where $N$ is the length of the mountain array.
Space Complexity: $O(1)$ .

Algorithm Strategy: Binary Search Patterns

Find the Peak Index:
- Use the logic from the "Find Min/Max" subpattern.
- Maintain low and high pointers.
- Compare get(mid) with get(mid + 1).
- If get(mid) < get(mid + 1), the peak is strictly to the right (low = mid + 1).
- Otherwise, the peak is at mid or to the left (high = mid).
- The loop terminates when low == high, giving us the peakIndex.
Search the Ascending Part:
- Perform a standard "Order-Agnostic" binary search on the range [0, peakIndex].
- Since this segment is sorted in ascending order, if get(mid) < target, we move right. If get(mid) > target, we move left.
- If found, return the index immediately (this guarantees the minimum index).
Search the Descending Part:
- If the target wasn't found in the ascending part, perform binary search on the range [peakIndex + 1, length - 1].
- Note that this segment is sorted in descending order.
- Logic is inverted: if get(mid) < target, we move left (towards larger values). If get(mid) > target, we move right.
- If found, return the index. Otherwise, return -1.

Execution Flow

Let's trace the algorithm with mountainArr = [1, 2, 3, 4, 5, 3, 1] and target = 3.

Step 1: Find Peak Index

Range [0, 6]. mid = 3.
Compare arr[3] (4) and arr[4] (5).
Since $4 < 5$ , we are ascending. Peak is to the right. New range [4, 6].
mid = 5. Compare arr[5] (3) and arr[6] (1).
Since $3 > 1$ , we are descending. Peak is at or left of 5. New range [4, 5].
mid = 4. Compare arr[4] (5) and arr[5] (3).
Since $5 > 3$ , we are descending. Peak is at or left of 4. New range [4, 4].
low == high. Peak Index is 4 (Value is 5).

Step 2: Search Left (Ascending) [0, 4]

Range [0, 4]. Target is 3.
mid = 2. arr[2] is 3.
Match found!
Since we want the minimum index and we search the left side first, we return 2.

Note: If the target were 3 in the right half only (e.g., if the array were [1, 5, 3]), Step 2 would fail, and we would proceed to Step 3.

Proof of Correctness

The correctness relies on the Bitonic Property of the Mountain Array.

Peak Uniqueness: A mountain array has exactly one peak. The condition arr[i] < arr[i+1] is true for all $i < \text{peak}$ and false for all $i \geq \text{peak}$ . This monotonicity in the slope allows binary search to converge on the peak index correctly.
Coverage: By splitting the array at the peak, we cover the entire search space. The left side [0, peak] is strictly increasing, and the right side [peak+1, end] is strictly decreasing.
Min Index Requirement: Searching the left subarray first ensures that if the target appears twice (once on each slope), we return the smaller index.

Pattern Reuse Notes

The logic used in Find in Mountain Array is a specific application of binary search on modified sorted arrays. This pattern recurs in several famous interview questions:

LeetCode 33: Search in Rotated Sorted Array: Similar to the mountain array, a rotated array consists of two sorted segments. You must determine which segment is sorted to decide pointer movement.
LeetCode 81: Search in Rotated Sorted Array II: Handles duplicates in the rotated array context, requiring worst-case $O(N)$ fallback but utilizing similar pivot logic.
LeetCode 153: Find Minimum in Rotated Sorted Array: This is the direct inverse of finding the peak. You use binary search to find the point where the order breaks (the minimum).
LeetCode 162: Find Peak Element: This is exactly the first step of the Mountain Array solution. It asks you to find any peak in an unsorted array where adjacent elements are never equal.

Feynman coach · Before you peek at the code

Can you explain the core insight in your own words?

If you can explain it before reading code, you’re far more likely to recall it under interview pressure.

Loading your attempt…

Reference Implementation

C++ Solution for LeetCode 1095

cpp

1/**
2 * // This is the MountainArray's API interface.
3 * // You should not implement it, or speculate about its implementation
4 * class MountainArray {
5 *   public:
6 *     int get(int index);
7 *     int length();
8 * };
9 */
10
11class Solution {
12public:
13    int findInMountainArray(int target, MountainArray &mountainArr) {
14        int n = mountainArr.length();
15        
16        // 1. Find the peak index
17        int low = 0;
18        int high = n - 1;
19        while (low < high) {
20            int mid = low + (high - low) / 2;
21            if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
22                // Ascending part, peak is to the right
23                low = mid + 1;
24            } else {
25                // Descending part or peak, peak is left or here
26                high = mid;
27            }
28        }
29        int peakIndex = low;
30        
31        // 2. Search in the ascending part (left of peak)
32        low = 0;
33        high = peakIndex;
34        while (low <= high) {
35            int mid = low + (high - low) / 2;
36            int val = mountainArr.get(mid);
37            if (val == target) return mid;
38            if (val < target) {
39                low = mid + 1;
40            } else {
41                high = mid - 1;
42            }
43        }
44        
45        // 3. Search in the descending part (right of peak)
46        low = peakIndex + 1;
47        high = n - 1;
48        while (low <= high) {
49            int mid = low + (high - low) / 2;
50            int val = mountainArr.get(mid);
51            if (val == target) return mid;
52            if (val < target) {
53                // In descending order, smaller value means we are too far right
54                high = mid - 1;
55            } else {
56                // Larger value means we are too far left
57                low = mid + 1;
58            }
59        }
60        
61        return -1;
62    }
63};

Java Solution for LeetCode 1095

java

1/**
2 * // This is MountainArray's API interface.
3 * // You should not implement it, or speculate about its implementation
4 * interface MountainArray {
5 *     public int get(int index);
6 *     public int length();
7 * }
8 */
9 
10class Solution {
11    public int findInMountainArray(int target, MountainArray mountainArr) {
12        int length = mountainArr.length();
13        
14        // 1. Find the peak index
15        int low = 0;
16        int high = length - 1;
17        while (low < high) {
18            int mid = low + (high - low) / 2;
19            if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
20                // We are in the increasing section
21                low = mid + 1;
22            } else {
23                // We are in the decreasing section or at the peak
24                high = mid;
25            }
26        }
27        int peakIndex = low;
28        
29        // 2. Binary Search on the increasing subarray [0, peakIndex]
30        int result = binarySearch(mountainArr, target, 0, peakIndex, true);
31        if (result != -1) {
32            return result;
33        }
34        
35        // 3. Binary Search on the decreasing subarray [peakIndex + 1, length - 1]
36        return binarySearch(mountainArr, target, peakIndex + 1, length - 1, false);
37    }
38    
39    private int binarySearch(MountainArray arr, int target, int low, int high, boolean ascending) {
40        while (low <= high) {
41            int mid = low + (high - low) / 2;
42            int val = arr.get(mid);
43            
44            if (val == target) {
45                return mid;
46            }
47            
48            if (ascending) {
49                if (val < target) {
50                    low = mid + 1;
51                } else {
52                    high = mid - 1;
53                }
54            } else {
55                // Descending logic
56                if (val < target) {
57                    high = mid - 1; // Target is larger, move left
58                } else {
59                    low = mid + 1;  // Target is smaller, move right
60                }
61            }
62        }
63        return -1;
64    }
65}

Python Solution for LeetCode 1095

python

1# """
2# This is MountainArray's API interface.
3# You should not implement it, or speculate about its implementation
4# """
5#class MountainArray:
6#    def get(self, index: int) -> int:
7#    def length(self) -> int:
8
9class Solution:
10    def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
11        n = mountain_arr.length()
12        
13        # 1. Find the peak index
14        low, high = 0, n - 1
15        while low < high:
16            mid = (low + high) // 2
17            if mountain_arr.get(mid) < mountain_arr.get(mid + 1):
18                low = mid + 1
19            else:
20                high = mid
21        peak_index = low
22        
23        # 2. Search in the ascending part
24        low, high = 0, peak_index
25        while low <= high:
26            mid = (low + high) // 2
27            val = mountain_arr.get(mid)
28            if val == target:
29                return mid
30            elif val < target:
31                low = mid + 1
32            else:
33                high = mid - 1
34                
35        # 3. Search in the descending part
36        low, high = peak_index + 1, n - 1
37        while low <= high:
38            mid = (low + high) // 2
39            val = mountain_arr.get(mid)
40            if val == target:
41                return mid
42            elif val < target:
43                # In descending, smaller value means go left (to larger values)
44                high = mid - 1
45            else:
46                low = mid + 1
47                
48        return -1

Find in Mountain Array

Core insight

Core Insight for Find in Mountain Array

Full breakdown

LeetCode 1095: Find in Mountain Array Solution & Explanation

Problem Overview

Brute Force Approach for Find in Mountain Array

Algorithm Strategy: Binary Search Patterns

Execution Flow

Proof of Correctness

Pattern Reuse Notes

Can you explain the core insight in your own words?

Reference Implementation

C++ Solution for LeetCode 1095

Java Solution for LeetCode 1095

Python Solution for LeetCode 1095

Practise Find in Mountain Array and the problems around it

Related problems

Find in Mountain Array

Core insight

Core Insight for Find in Mountain Array

Full breakdown

LeetCode 1095: Find in Mountain Array Solution & Explanation

Problem Overview

Brute Force Approach for Find in Mountain Array

Algorithm Strategy: Binary Search Patterns

Execution Flow

Proof of Correctness

Pattern Reuse Notes

Can you explain the core insight in your own words?

Reference Implementation

C++ Solution for LeetCode 1095

Java Solution for LeetCode 1095

Python Solution for LeetCode 1095

Practise Find in Mountain Array and the problems around it

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