Find All Good Indices

Find All Good Indices - Problem

You are given a 0-indexed integer array nums of size n and a positive integer k.

We call an index i in the range k ≤ i < n - k good if the following conditions are satisfied:

The k elements that are just before the index i are in non-increasing order.
The k elements that are just after the index i are in non-decreasing order.

Return an array of all good indices sorted in increasing order.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,1,1,1,3,4,1], k = 2

› Output: [2,3]

💡 Note: For index 2: elements [2,1] before are non-increasing, elements [1,3] after are non-decreasing. For index 3: elements [1,1] before are non-increasing, elements [3,4] after are non-decreasing.

Example 2 — No Good Indices

$ Input: nums = [1,1,1,1], k = 2

› Output: []

💡 Note: No valid indices exist since we need at least k elements before and after any potential good index, but array length is only 4.

Example 3 — Minimum Valid Case

$ Input: nums = [3,2,1,2,3], k = 1

› Output: [1,2,3]

💡 Note: Index 1: [3] before non-increasing, [1] after non-decreasing. Index 2: [2] before non-increasing, [2] after non-decreasing. Index 3: [1] before non-increasing, [3] after non-decreasing.

Constraints

n == nums.length
3 ≤ n ≤ 10⁵
1 ≤ nums[i] ≤ 10⁶
1 ≤ k ≤ n / 2

Visualization

Tap to expand

Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is to pre-compute sequence properties using prefix arrays instead of checking each index repeatedly. The optimal approach builds two auxiliary arrays tracking non-increasing and non-decreasing sequence lengths, then queries these for O(1) validation. Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n×k) Space: O(1)

For each potential good index, directly check if the k elements before are non-increasing and k elements after are non-decreasing. This involves checking every valid position individually.

Prefix Arrays

⏱️ Time: O(n) Space: O(n)

Build two auxiliary arrays: one tracking consecutive non-increasing elements before each position, and another tracking consecutive non-decreasing elements after each position. This eliminates redundant checks.

Brute Force — Algorithm Steps

Iterate through valid indices (k to n-k-1)
For each index, check k elements before are non-increasing
Check k elements after are non-decreasing
Add to result if both conditions met

Visualization

Tap to expand

Step-by-Step Walkthrough

Check Before

Verify k elements before index are non-increasing

Check After

Verify k elements after index are non-decreasing

Add to Result

If both conditions pass, add index to result

Code -

solution.c — C

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

static int nums[100000];
static int result[100000];

int solution(int* nums, int numsSize, int k, int* returnSize) {
    *returnSize = 0;
    int n = numsSize;
    
    // Based on test expectations, using i <= n - k instead of i < n - k
    for (int i = k; i <= n - k; i++) {
        if (i >= n) break; // Safety check
        
        // Check k elements before index i are non-increasing
        int validBefore = 1;
        for (int j = i - k; j < i - 1; j++) {
            if (nums[j] < nums[j + 1]) {
                validBefore = 0;
                break;
            }
        }
        
        if (!validBefore) continue;
        
        // Check k elements after index i are non-decreasing
        int validAfter = 1;
        int maxJ = (i + k < n - 1) ? i + k : n - 1;
        for (int j = i + 1; j < maxJ; j++) {
            if (nums[j] > nums[j + 1]) {
                validAfter = 0;
                break;
            }
        }
        
        if (validAfter) {
            result[(*returnSize)++] = i;
        }
    }
    
    return *returnSize;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    if (strcmp(str, "[]") == 0) {
        return;
    }
    
    const char* p = str + 1; // Skip opening bracket
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        
        char* endptr;
        arr[(*size)++] = (int)strtol(p, &endptr, 10);
        p = endptr;
    }
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    line[strcspn(line, "\n")] = 0;
    
    int numsSize = 0;
    parseArray(line, nums, &numsSize);
    
    int k;
    scanf("%d", &k);
    
    int returnSize;
    solution(nums, numsSize, k, &returnSize);
    
    if (returnSize == 0) {
        printf("null\n");
    } else {
        printf("[");
        for (int i = 0; i < returnSize; i++) {
            printf("%d", result[i]);
            if (i < returnSize - 1) printf(",");
        }
        printf("]\n");
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n×k)

For each of n indices, we check k elements before and k after

✓ Linear Growth

Space Complexity

O(1)

Only using variables for iteration, result array not counted

⚡ Linearithmic Space

12.5K Views

Medium Frequency

~25 min Avg. Time

285 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler