LeetCode #1095 — HARD

Find in Mountain Array

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

(This problem is an interactive problem.)

You may recall that an array arr is a mountain array if and only if:

arr.length >= 3
There exists some i with 0 < i < arr.length - 1 such that:
- arr[0] < arr[1] < ... < arr[i - 1] < arr[i]
- arr[i] > arr[i + 1] > ... > arr[arr.length - 1]

Given a mountain array mountainArr, return the minimum index such that mountainArr.get(index) == target. If such an index does not exist, return -1.

You cannot access the mountain array directly. You may only access the array using a MountainArray interface:

MountainArray.get(k) returns the element of the array at index k (0-indexed).
MountainArray.length() returns the length of the array.

Submissions making more than 100 calls to MountainArray.get will be judged Wrong Answer. Also, any solutions that attempt to circumvent the judge will result in disqualification.

Example 1:

Input: mountainArr = [1,2,3,4,5,3,1], target = 3
Output: 2
Explanation: 3 exists in the array, at index=2 and index=5. Return the minimum index, which is 2.

Example 2:

Input: mountainArr = [0,1,2,4,2,1], target = 3
Output: -1
Explanation: 3 does not exist in the array, so we return -1.

Constraints:

3 <= mountainArr.length() <= 10⁴
0 <= target <= 10⁹
0 <= mountainArr.get(index) <= 10⁹

Step 01

Brute Force Baseline

Problem summary: (This problem is an interactive problem.) You may recall that an array arr is a mountain array if and only if: arr.length >= 3 There exists some i with 0 < i < arr.length - 1 such that: arr[0] < arr[1] < ... < arr[i - 1] < arr[i] arr[i] > arr[i + 1] > ... > arr[arr.length - 1] Given a mountain array mountainArr, return the minimum index such that mountainArr.get(index) == target. If such an index does not exist, return -1. You cannot access the mountain array directly. You may only access the array using a MountainArray interface: MountainArray.get(k) returns the element of the array at index k (0-indexed). MountainArray.length() returns the length of the array. Submissions making more than 100 calls to MountainArray.get will be judged Wrong Answer. Also, any solutions that attempt to circumvent the judge will result in disqualification.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[1,2,3,4,5,3,1]
3

Example 2

[0,1,2,4,2,1]
3

Core Insight

What unlocks the optimal approach

Based on whether A[i-1] < A[i] < A[i+1], A[i-1] < A[i] > A[i+1], or A[i-1] > A[i] > A[i+1], we are either at the left side, peak, or right side of the mountain. We can binary search to find the peak. After finding the peak, we can binary search two more times to find whether the value occurs on either side of the peak.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1095: Find in Mountain Array
/**
 * // This is MountainArray's API interface.
 * // You should not implement it, or speculate about its implementation
 * interface MountainArray {
 *     public int get(int index) {}
 *     public int length() {}
 * }
 */

class Solution {
    private MountainArray mountainArr;
    private int target;

    public int findInMountainArray(int target, MountainArray mountainArr) {
        int n = mountainArr.length();
        int l = 0, r = n - 1;
        while (l < r) {
            int mid = (l + r) >>> 1;
            if (mountainArr.get(mid) > mountainArr.get(mid + 1)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        this.mountainArr = mountainArr;
        this.target = target;
        int ans = search(0, l, 1);
        return ans == -1 ? search(l + 1, n - 1, -1) : ans;
    }

    private int search(int l, int r, int k) {
        while (l < r) {
            int mid = (l + r) >>> 1;
            if (k * mountainArr.get(mid) >= k * target) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return mountainArr.get(l) == target ? l : -1;
    }
}

// Accepted solution for LeetCode #1095: Find in Mountain Array
/**
 * // This is the MountainArray's API interface.
 * // You should not implement it, or speculate about its implementation
 * type MountainArray struct {
 * }
 *
 * func (this *MountainArray) get(index int) int {}
 * func (this *MountainArray) length() int {}
 */

func findInMountainArray(target int, mountainArr *MountainArray) int {
	n := mountainArr.length()
	l, r := 0, n-1
	for l < r {
		mid := (l + r) >> 1
		if mountainArr.get(mid) > mountainArr.get(mid+1) {
			r = mid
		} else {
			l = mid + 1
		}
	}
	search := func(l, r, k int) int {
		for l < r {
			mid := (l + r) >> 1
			if k*mountainArr.get(mid) >= k*target {
				r = mid
			} else {
				l = mid + 1
			}
		}
		if mountainArr.get(l) == target {
			return l
		}
		return -1
	}
	ans := search(0, l, 1)
	if ans == -1 {
		return search(l+1, n-1, -1)
	}
	return ans
}

# Accepted solution for LeetCode #1095: Find in Mountain Array
# """
# This is MountainArray's API interface.
# You should not implement it, or speculate about its implementation
# """
# class MountainArray:
#    def get(self, index: int) -> int:
#    def length(self) -> int:


class Solution:
    def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
        def search(l: int, r: int, k: int) -> int:
            while l < r:
                mid = (l + r) >> 1
                if k * mountain_arr.get(mid) >= k * target:
                    r = mid
                else:
                    l = mid + 1
            return -1 if mountain_arr.get(l) != target else l

        n = mountain_arr.length()
        l, r = 0, n - 1
        while l < r:
            mid = (l + r) >> 1
            if mountain_arr.get(mid) > mountain_arr.get(mid + 1):
                r = mid
            else:
                l = mid + 1
        ans = search(0, l, 1)
        return search(l + 1, n - 1, -1) if ans == -1 else ans

// Accepted solution for LeetCode #1095: Find in Mountain Array
impl Solution {
    #[allow(dead_code)]
    pub fn find_in_mountain_array(target: i32, mountain_arr: &MountainArray) -> i32 {
        let n = mountain_arr.length();

        // First find the maximum element in the array
        let mut l = 0;
        let mut r = n - 1;

        while l < r {
            let mid = (l + r) >> 1;
            if mountain_arr.get(mid) > mountain_arr.get(mid + 1) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }

        let left = Self::binary_search(mountain_arr, 0, l, 1, target);

        if left == -1 {
            Self::binary_search(mountain_arr, l, n - 1, -1, target)
        } else {
            left
        }
    }

    #[allow(dead_code)]
    fn binary_search(m: &MountainArray, mut l: i32, mut r: i32, k: i32, target: i32) -> i32 {
        let n = m.length();

        while l < r {
            let mid = (l + r) >> 1;
            if k * m.get(mid) >= k * target {
                r = mid;
            } else {
                l = mid + 1;
            }
        }

        if m.get(l) == target {
            l
        } else {
            -1
        }
    }
}

// Accepted solution for LeetCode #1095: Find in Mountain Array
/**
 * // This is the MountainArray's API interface.
 * // You should not implement it, or speculate about its implementation
 * class Master {
 *      get(index: number): number {}
 *
 *      length(): number {}
 * }
 */

function findInMountainArray(target: number, mountainArr: MountainArray) {
    const n = mountainArr.length();
    let l = 0;
    let r = n - 1;
    while (l < r) {
        const mid = (l + r) >> 1;
        if (mountainArr.get(mid) > mountainArr.get(mid + 1)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    const search = (l: number, r: number, k: number): number => {
        while (l < r) {
            const mid = (l + r) >> 1;
            if (k * mountainArr.get(mid) >= k * target) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return mountainArr.get(l) === target ? l : -1;
    };
    const ans = search(0, l, 1);
    return ans === -1 ? search(l + 1, n - 1, -1) : ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.