1200. Minimum Absolute Difference

Given an array of distinct integers arr, find all pairs of elements with the minimum absolute difference of any two elements.

Return a list of pairs in ascending order(with respect to pairs), each pair [a, b] follows

• a, b are from arr
• a < b
• b - a equals to the minimum absolute difference of any two elements in arr
   

Example 1:

Input: arr = [4,2,1,3]
Output: [[1,2],[2,3],[3,4]]
Explanation: The minimum absolute difference is 1. List all pairs with difference equal to 1 in ascending order.

Example 2:

Input: arr = [1,3,6,10,15]
Output: [[1,3]]

Example 3:

Input: arr = [3,8,-10,23,19,-4,-14,27]
Output: [[-14,-10],[19,23],[23,27]]

Constraints:

• $2<=arr.length<=10^5$
• $-10^6<=arr[i]<=10^6$

From: LeetCode
Link: 1200. Minimum Absolute Difference

Solution:

Ideas:

Sort first, then only compare neighboring numbers.

Code:

#include <stdlib.h>

int cmpInt(const void* a, const void* b) {
    int x = *(int*)a;
    int y = *(int*)b;
    return (x > y) - (x < y);
}

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** minimumAbsDifference(int* arr, int arrSize, int* returnSize, int** returnColumnSizes) {
    qsort(arr, arrSize, sizeof(int), cmpInt);

    int minDiff = arr[1] - arr[0];
    for (int i = 1; i < arrSize - 1; i++) {
        int diff = arr[i + 1] - arr[i];
        if (diff < minDiff) {
            minDiff = diff;
        }
    }

    int count = 0;
    for (int i = 0; i < arrSize - 1; i++) {
        if (arr[i + 1] - arr[i] == minDiff) {
            count++;
        }
    }

    int** result = (int**)malloc(sizeof(int*) * count);
    *returnColumnSizes = (int*)malloc(sizeof(int) * count);
    *returnSize = count;

    int idx = 0;
    for (int i = 0; i < arrSize - 1; i++) {
        if (arr[i + 1] - arr[i] == minDiff) {
            result[idx] = (int*)malloc(sizeof(int) * 2);
            result[idx][0] = arr[i];
            result[idx][1] = arr[i + 1];
            (*returnColumnSizes)[idx] = 2;
            idx++;
        }
    }

    return result;
}