Minimum Threshold for Inversion Pairs Count

Minimum Threshold for Inversion Pairs Count - Problem

You are given an array of integers nums and an integer k. An inversion pair with a threshold x is defined as a pair of indices (i, j) such that:

i < j
nums[i] > nums[j]
The difference between the two numbers is at most x (i.e. nums[i] - nums[j] <= x)

Your task is to determine the minimum integer min_threshold such that there are at least k inversion pairs with threshold min_threshold. If no such integer exists, return -1.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,3,6,2], k = 2

› Output: 1

💡 Note: Inversion pairs: (0,1) with diff=1, (0,3) with diff=2, (1,3) with diff=1, (2,3) with diff=4. With threshold=1, we have pairs (0,1) and (1,3), which gives us 2 pairs ≥ k=2.

Example 2 — Need Higher Threshold

$ Input: nums = [5,4,3,2], k = 3

› Output: 2

💡 Note: Inversion pairs: (0,1)→1, (0,2)→2, (0,3)→3, (1,2)→1, (1,3)→2, (2,3)→1. Sorted differences: [1,1,1,2,2,3]. The 3rd smallest is 1, but we need threshold=2 to get exactly 3 pairs with diff ≤ 2.

Example 3 — No Solution

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

› Output: -1

💡 Note: Array is already sorted, so no inversion pairs exist. Cannot find k=1 pairs, return -1.

Constraints

2 ≤ nums.length ≤ 500
1 ≤ nums[i] ≤ 10⁶
0 ≤ k ≤ n(n-1)/2

Visualization

Tap to expand

Understanding the Visualization

Input

Array [4,3,6,2] and k=2 (need 2+ pairs)

Find Pairs

Identify inversion pairs and their differences

Minimum Threshold

Find smallest threshold giving ≥k pairs

Key Takeaway

🎯 Key Insight: The minimum threshold is the k-th smallest difference among all inversion pairs

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that we need to find the k-th smallest difference among all inversion pairs. Best approach is to collect all inversion pair differences, sort them, and return the k-th element. Time: O(n² log n), Space: O(n²)

Common Approaches

Approach	Time	Space	Notes
✓ Binary Search on Answer	O(n² log n)	O(n²)	Binary search on threshold value combined with efficient pair counting
Brute Force	O(n² × max_diff)	O(1)	Try all possible thresholds from 0 to maximum difference
Sort and Count Efficiently	O(n² + n² log n)	O(n²)	Sort differences once, then use binary search or direct indexing to find answer

Binary Search on Answer — Algorithm Steps

Collect all possible differences from inversion pairs
Binary search on sorted differences
For each mid value, count pairs with difference ≤ mid
Adjust search range based on count vs k

Visualization

Tap to expand

Step-by-Step Walkthrough

Collect Differences

Find all inversion pair differences and sort them

Binary Search

Search threshold space using binary search

Count & Adjust

Count pairs for each threshold and adjust search range

Code -

solution.c — C

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

int compare(const void* a, const void* b) {
    return (*(int*)a - *(int*)b);
}

int solution(int* nums, int numsSize, int k) {
    if (k == 0) return 0;
    
    // Find all inversion pairs and their differences
    int* differences = (int*)malloc(numsSize * numsSize * sizeof(int));
    int diffCount = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (nums[i] > nums[j]) {
                differences[diffCount++] = nums[i] - nums[j];
            }
        }
    }
    
    if (diffCount < k) {
        free(differences);
        return -1;
    }
    
    // Sort differences for binary search
    qsort(differences, diffCount, sizeof(int), compare);
    
    // Binary search on the answer
    int left = 0, right = differences[diffCount - 1];
    int result = -1;
    
    while (left <= right) {
        int mid = left + (right - left) / 2;
        int count = 0;
        
        // Count pairs with difference <= mid
        for (int i = 0; i < diffCount; i++) {
            if (differences[i] <= mid) count++;
            else break;
        }
        
        if (count >= k) {
            result = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    
    free(differences);
    return result;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n² log n)

O(n²) to find all inversion pairs, O(log n) binary search iterations, O(n²) counting per iteration

⚠ Quadratic Growth

Space Complexity

O(n²)

Store all inversion pair differences in array

⚠ Quadratic Space

8.9K Views

Medium Frequency

~25 min Avg. Time

234 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

Binary Search on Answer — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler