Count Pairs That Form a Complete Day I

Count Pairs That Form a Complete Day I - Problem

Given an integer array hours representing times in hours, return an integer denoting the number of pairs i, j where i < j and hours[i] + hours[j] forms a complete day.

A complete day is defined as a time duration that is an exact multiple of 24 hours. For example, 1 day is 24 hours, 2 days is 48 hours, 3 days is 72 hours, and so on.

Input & Output

Example 1 — Basic Case

$ Input: hours = [12,12,30,24,24]

› Output: 2

💡 Note: Valid pairs: (0,1) → 12+12=24, (3,4) → 24+24=48. Both sums are multiples of 24.

Example 2 — No Valid Pairs

$ Input: hours = [72,48,24,3]

› Output: 3

💡 Note: Valid pairs: (0,1) → 72+48=120, (0,2) → 72+24=96, (1,2) → 48+24=72. All are multiples of 24.

Example 3 — Single Element

$ Input: hours = [1,2]

› Output: 0

💡 Note: 1+2=3 is not a multiple of 24, so no valid pairs exist.

Constraints

1 ≤ hours.length ≤ 100
1 ≤ hours[i] ≤ 10⁹

Visualization

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Asked in

a Amazon 35 M Microsoft 28 G Google 22

The key insight is to use modulo arithmetic: two hours form a complete day if (hour1 + hour2) % 24 = 0. Best approach is frequency counting with O(n) time and O(1) space by grouping hours by their remainder when divided by 24.

Common Approaches

✓ Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

Use nested loops to examine all possible pairs (i, j) where i < j, and count pairs where hours[i] + hours[j] is divisible by 24.

Frequency Count with Modulo

⏱️ Time: O(n) Space: O(1)

Calculate hour % 24 for each element to find remainders, then count pairs where remainders sum to 0 or 24. Use frequency map to avoid nested loops.

Brute Force - Check All Pairs — Algorithm Steps

Iterate through each index i from 0 to n-2
For each i, iterate j from i+1 to n-1
Check if (hours[i] + hours[j]) % 24 == 0
Count valid pairs

Visualization

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Step-by-Step Walkthrough

Outer Loop

Fix first element at index i

Inner Loop

Check all j > i for valid pairs

Count

Increment count when sum % 24 = 0

Code -

solution.c — C

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

int solution(int* hours, int hoursSize) {
    int count = 0;
    
    for (int i = 0; i < hoursSize; i++) {
        for (int j = i + 1; j < hoursSize; j++) {
            if ((hours[i] + hours[j]) % 24 == 0) {
                count++;
            }
        }
    }
    
    return count;
}

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

✓ Linear Space

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Medium Frequency

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Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

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Code -

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