Find All Good Indices - Practice Coding Problems

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

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Understanding the Visualization

Input Array

Array with elements and target k value

Check Conditions

Verify k elements before non-increasing, k after non-decreasing

Good Indices

Return all indices satisfying both conditions

Key Takeaway

🎯 Key Insight: Pre-compute sequence properties to avoid redundant validation checks

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

Approach	Time	Space	Notes
✓ Prefix Arrays	O(n)	O(n)	Pre-compute non-increasing and non-decreasing sequence lengths
Brute Force	O(n×k)	O(1)	Check each valid index by examining k elements before and after

Prefix Arrays — Algorithm Steps

Build prefix array for non-increasing sequences
Build suffix array for non-decreasing sequences
Check each valid index using precomputed values
Add indices where both conditions are satisfied

Visualization

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Step-by-Step Walkthrough

Build Prefix

Count consecutive non-increasing elements ending at each position

Build Suffix

Count consecutive non-decreasing elements starting at each position

Query Results

Use precomputed values to find good indices

Code -

solution.c — C

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

int* solution(int* nums, int numsSize, int k, int* returnSize) {
    if (numsSize < 2 * k + 1) {
        *returnSize = 0;
        return NULL;
    }
    
    // Build prefix array: nonIncreasing[i] = length of non-increasing sequence ending at i
    int* nonIncreasing = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        nonIncreasing[i] = 1;
    }
    for (int i = 1; i < numsSize; i++) {
        if (nums[i] <= nums[i - 1]) {
            nonIncreasing[i] = nonIncreasing[i - 1] + 1;
        }
    }
    
    // Build suffix array: nonDecreasing[i] = length of non-decreasing sequence starting at i
    int* nonDecreasing = (int*)malloc(numsSize * sizeof(int));
    for (int i = 0; i < numsSize; i++) {
        nonDecreasing[i] = 1;
    }
    for (int i = numsSize - 2; i >= 0; i--) {
        if (nums[i] <= nums[i + 1]) {
            nonDecreasing[i] = nonDecreasing[i + 1] + 1;
        }
    }
    
    int* result = (int*)malloc(numsSize * sizeof(int));
    int count = 0;
    
    for (int i = k; i < numsSize - k; i++) {
        if (nonIncreasing[i - 1] >= k && nonDecreasing[i + 1] >= k) {
            result[count++] = i;
        }
    }
    
    *returnSize = count;
    free(nonIncreasing);
    free(nonDecreasing);
    return result;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        int num = atoi(token);
        nums[numsSize++] = num;
        token = strtok(NULL, "[,]");
    }
    
    int k;
    scanf("%d", &k);
    
    int returnSize;
    int* result = solution(nums, numsSize, k, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    if (result) free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Three linear passes: build prefix array, build suffix array, check results

✓ Linear Growth

Space Complexity

O(n)

Two auxiliary arrays of size n for prefix and suffix information

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Prefix Arrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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