Minimum Absolute Difference Between Elements With Constraint

Minimum Absolute Difference Between Elements With Constraint - Problem

You are given a 0-indexed integer array nums and an integer x.

Find the minimum absolute difference between two elements in the array that are at least x indices apart.

In other words, find two indices i and j such that abs(i - j) >= x and abs(nums[i] - nums[j]) is minimized.

Return an integer denoting the minimum absolute difference between two elements that are at least x indices apart.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,3,2,4], x = 2

› Output: 0

💡 Note: We can choose indices i=0 and j=3 where abs(0-3) = 3 >= 2 and abs(nums[0] - nums[3]) = abs(4-4) = 0

Example 2 — No Exact Match

$ Input: nums = [5,3,2,10,15], x = 1

› Output: 1

💡 Note: The minimum difference is 1 between elements 3 and 2 at indices 1 and 2 where abs(1-2) = 1 >= 1

Example 3 — Larger Gap Required

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

› Output: 3

💡 Note: Only valid pair is indices 0 and 3: abs(0-3) = 3 >= 3 and abs(1-4) = 3

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ x ≤ nums.length - 1
0 ≤ nums[i] ≤ 10⁹

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is to use a sorted data structure to efficiently find the closest value among all valid candidates. The optimal approach maintains a sorted set of elements that are at least x positions behind the current position. Time: O(n log n), Space: O(n)

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each element, check all previous elements that are at least x positions away and find the minimum absolute difference. This ensures we consider all valid pairs.

Sliding Window with Sorted Set

⏱️ Time: O(n log n) Space: O(n)

As we iterate through the array, maintain a sorted set of all elements that are at least x positions behind. For each new element, find the closest value in the sorted set using binary search.

Brute Force — Algorithm Steps

Step 1: For each index i from x to n-1
Step 2: Check all indices j from 0 to i-x
Step 3: Calculate abs(nums[i] - nums[j]) and track minimum

Visualization

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Step-by-Step Walkthrough

Start at index x

Begin checking from position x onwards

Compare with all valid previous

For each position, check all elements at least x positions back

Track minimum

Keep track of the smallest absolute difference found

Code -

solution.c — C

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

int abs_val(int a) {
    return a < 0 ? -a : a;
}

int min_val(int a, int b) {
    return a < b ? a : b;
}

int solution(int* nums, int numsSize, int x) {
    int minDiff = INT_MAX;
    
    for (int i = x; i < numsSize; i++) {
        for (int j = 0; j <= i - x; j++) {
            int diff = abs_val(nums[i] - nums[j]);
            minDiff = min_val(minDiff, diff);
        }
    }
    
    return minDiff;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int numsSize = 0;
    parseArray(line, nums, &numsSize);
    
    int x;
    scanf("%d", &x);
    
    int result = solution(nums, numsSize, x);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops where outer loop runs n-x times and inner loop runs up to n-x times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables to track minimum difference

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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