Count Number of Bad Pairs - Practice Coding Problems

Count Number of Bad Pairs - Problem

You are given a 0-indexed integer array nums. A pair of indices (i, j) is a bad pair if i < j and j - i != nums[j] - nums[i].

Return the total number of bad pairs in nums.

Input & Output

Example 1 — Basic Case

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

› Output: 5

💡 Note: The bad pairs are (0,1): 1-0=1 ≠ 1-4=-3, (0,2): 2-0=2 ≠ 3-4=-1, (0,3): 3-0=3 ≠ 3-4=-1, (1,2): 2-1=1 ≠ 3-1=2, (2,3): 3-2=1 ≠ 3-3=0. Total = 5 bad pairs.

Example 2 — Small Array

$ Input: nums = [1,2]

› Output: 0

💡 Note: Only one pair (0,1): j-i = 1-0 = 1, nums[j]-nums[i] = 2-1 = 1. Since 1 = 1, this is a good pair, so 0 bad pairs.

Example 3 — All Bad Pairs

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

› Output: 0

💡 Note: All pairs have the same value: (0,1): 1-0=1, 1-1=0 (1≠0), (0,2): 2-0=2, 1-1=0 (2≠0), (1,2): 2-1=1, 1-1=0 (1≠0). Wait, let me recalculate: for identical elements, nums[j]-nums[i]=0, but j-i>0, so all pairs are bad. Total = 3 bad pairs.

Constraints

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

Visualization

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Understanding the Visualization

Input Array

Array of integers with indices

Check Pairs

For each pair (i,j), check if j-i ≠ nums[j]-nums[i]

Count Bad

Count pairs that satisfy the bad condition

Key Takeaway

🎯 Key Insight: Transform the condition to count good pairs efficiently, then subtract from total pairs

Asked in

G Google 25 f Facebook 20 a Amazon 15

The key insight is transforming the bad pair condition j - i != nums[j] - nums[i] into a good pair condition by rearranging to nums[i] - i = nums[j] - j. We can count good pairs efficiently using frequency counting, then subtract from total pairs. Best approach uses frequency counting with Time: O(n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Pairs	O(n²)	O(1)	Check every possible pair (i,j) and count those that don't satisfy the condition
Frequency Count - Count Good Pairs	O(n)	O(n)	Count good pairs using frequency map, then subtract from total pairs

Brute Force - Check All Pairs — Algorithm Steps

Step 1: Use nested loops to generate all pairs (i,j) where i < j
Step 2: For each pair, check if j - i != nums[j] - nums[i]
Step 3: Count pairs that satisfy the bad pair condition

Visualization

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Step-by-Step Walkthrough

Generate Pairs

Create all pairs (i,j) where i < j

Check Condition

For each pair, check if j - i != nums[j] - nums[i]

Count Bad Pairs

Increment counter for pairs that satisfy the condition

Code -

solution.c — C

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

long long solution(int* nums, int numsSize) {
    long long badPairs = 0;
    
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (j - i != nums[j] - nums[i]) {
                badPairs++;
            }
        }
    }
    
    return badPairs;
}

int main() {
    char line[100000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[10000];
    int size = 0;
    
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != '\n') {
            nums[size++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    long long result = solution(nums, size);
    printf("%lld\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops check all n(n-1)/2 pairs

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses constant extra variables for counting

✓ Linear Space

28.0K Views

Medium Frequency

~25 min Avg. Time

890 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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