Find Indices of Stable Mountains - Problem

Array Easy

There are n mountains in a row, and each mountain has a height. You are given an integer array height where height[i] represents the height of mountain i, and an integer threshold.

A mountain is called stable if the mountain just before it (if it exists) has a height strictly greater than threshold.

Note: Mountain 0 is not stable.

Return an array containing the indices of all stable mountains in any order.

Input & Output

Example 1 — Basic Case

$ Input: height = [1,2,3,4,5], threshold = 2

› Output: [3,4]

💡 Note: Mountain 3 is stable because height[2] = 3 > 2. Mountain 4 is stable because height[3] = 4 > 2. Mountains 0,1,2 are not stable.

Example 2 — No Stable Mountains

$ Input: height = [2,1,1], threshold = 5

› Output: []

💡 Note: No mountain is stable because height[0] = 2 ≤ 5 and height[1] = 1 ≤ 5.

Example 3 — All Mountains Stable

$ Input: height = [10,1,10,1,10], threshold = 3

› Output: [1,3]

💡 Note: Mountains 1 and 3 are stable because height[0] = 10 > 3 and height[2] = 10 > 3. Mountains 2 and 4 are not stable because height[1] = 1 ≤ 3 and height[3] = 1 ≤ 3.

Constraints

1 ≤ height.length ≤ 100
1 ≤ height[i] ≤ 100
1 ≤ threshold ≤ 100

Visualization

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Asked in

M Microsoft 15 A Apple 12

This problem requires a single pass through the mountain array. The key insight is that we only need to check if each mountain's predecessor exceeds the threshold. Best approach is single iteration starting from index 1. Time: O(n), Space: O(1).

Common Approaches

✓ Bfs

⏱️ Time: N/A Space: N/A

Single Pass Check

⏱️ Time: O(n) Space: O(1)

Start from index 1 (since mountain 0 cannot be stable) and for each mountain, check if the previous mountain's height is strictly greater than the threshold. If so, add the current index to the result.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

static int result[1000];
static int heights[1000];

int parseArray(char* line, int* arr) {
    int count = 0;
    int i = 0;
    int len = strlen(line);
    
    // Skip whitespace and find '['
    while (i < len && line[i] != '[') i++;
    if (i >= len) return 0;
    i++; // skip '['
    
    // Handle empty array
    while (i < len && (line[i] == ' ' || line[i] == '\t')) i++;
    if (i < len && line[i] == ']') return 0;
    
    while (i < len) {
        // Skip whitespace and commas
        while (i < len && (line[i] == ' ' || line[i] == '\t' || line[i] == ',')) i++;
        
        if (i >= len || line[i] == ']') break;
        
        // Parse number
        int num = 0;
        int sign = 1;
        if (line[i] == '-') {
            sign = -1;
            i++;
        }
        
        while (i < len && line[i] >= '0' && line[i] <= '9') {
            num = num * 10 + (line[i] - '0');
            i++;
        }
        
        arr[count++] = sign * num;
    }
    
    return count;
}

int* stableMountains(int* height, int heightSize, int threshold, int* returnSize) {
    int count = 0;
    
    // Mountain 0 is never stable
    for (int i = 1; i < heightSize; i++) {
        if (height[i-1] > threshold) {
            result[count++] = i;
        }
    }
    
    *returnSize = count;
    return result;
}

int main() {
    char heightStr[10000];
    char thresholdStr[100];
    
    fgets(heightStr, sizeof(heightStr), stdin);
    fgets(thresholdStr, sizeof(thresholdStr), stdin);
    
    int heightSize = parseArray(heightStr, heights);
    int threshold = atoi(thresholdStr);
    
    int returnSize;
    int* stableIndices = stableMountains(heights, heightSize, threshold, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("%d", stableIndices[i]);
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~10 min Avg. Time

245 Likes

Ln 1, Col 1

Smart Actions

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