Number of Excellent Pairs - Problem

You are given a 0-indexed positive integer array nums and a positive integer k.

A pair of numbers (num1, num2) is called excellent if the following conditions are satisfied:

Both the numbers num1 and num2 exist in the array nums.
The sum of the number of set bits in num1 OR num2 and num1 AND num2 is greater than or equal to k, where OR is the bitwise OR operation and AND is the bitwise AND operation.

Return the number of distinct excellent pairs.

Two pairs (a, b) and (c, d) are considered distinct if either a != c or b != d. For example, (1, 2) and (2, 1) are distinct.

Note that a pair (num1, num2) such that num1 == num2 can also be excellent if you have at least one occurrence of num1 in the array.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,5], k = 5

› Output: 0

💡 Note: No pairs have sufficient set bits. For any pair (a,b) from [1,2,3,5], the maximum sum of set bits in (a|b) and (a&b) is 4, which is less than k=5.

Example 2 — All Pairs Valid

$ Input: nums = [5,1,3], k = 3

› Output: 8

💡 Note: All pairs except (1,1) are excellent. 1 has 1 bit, 3 has 2 bits, 5 has 2 bits. Valid pairs: (1,3)→3, (1,5)→3, (3,1)→3, (3,3)→4, (3,5)→4, (5,1)→3, (5,3)→4, (5,5)→4. Total: 8 pairs.

Example 3 — No Valid Pairs

$ Input: nums = [1,2], k = 5

› Output: 0

💡 Note: Maximum bit sum is 1+1=2 for (1,2), which is less than k=5. No excellent pairs exist.

Constraints

1 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
1 ≤ k ≤ 60

Visualization

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Asked in

G Google 12 M Meta 8 M Microsoft 6

The key insight is using the mathematical property that bits(a|b) + bits(a&b) = bits(a) + bits(b). Instead of computing OR and AND for each pair, group numbers by bit count and multiply combinations. Best approach is bit count grouping with Time: O(n + B²), Space: O(B).

Common Approaches

✓ Bit Count Grouping

⏱️ Time: O(n + B²) Space: O(B)

Group numbers by their bit count, then use the key insight that for any two numbers a and b, the sum of set bits in (a OR b) and (a AND b) equals the sum of set bits in a and b individually.

Brute Force

⏱️ Time: O(n² × log(max(nums))) Space: O(n)

For each pair of numbers in the array, calculate the bitwise OR and AND, count the set bits in both results, and check if their sum meets the threshold k.

Bit Count Grouping — Algorithm Steps

Remove duplicates and count bit counts
Group numbers by their bit count
For each pair of bit counts, check if sum >= k
Count all valid combinations

Visualization

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Step-by-Step Walkthrough

Count Bits

Group numbers by their bit count

Mathematical Property

bits(a|b) + bits(a&b) = bits(a) + bits(b)

Count Combinations

Multiply counts for valid bit sum pairs

Code -

solution.c — C

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

int countBits(int x) {
    int count = 0;
    while (x) {
        count += x & 1;
        x >>= 1;
    }
    return count;
}

long long solution(int* nums, int numsSize, int k) {
    int unique[1000];
    int uniqueSize = 0;
    
    // Remove duplicates
    for (int i = 0; i < numsSize; i++) {
        int found = 0;
        for (int j = 0; j < uniqueSize; j++) {
            if (unique[j] == nums[i]) {
                found = 1;
                break;
            }
        }
        if (!found) {
            unique[uniqueSize++] = nums[i];
        }
    }
    
    long long count = 0;
    for (int i = 0; i < uniqueSize; i++) {
        for (int j = 0; j < uniqueSize; j++) {
            int num1 = unique[i], num2 = unique[j];
            int orBits = countBits(num1 | num2);
            int andBits = countBits(num1 & num2);
            if (orBits + andBits >= k) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[numsSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    printf("%lld\n", solution(nums, numsSize, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + B²)

O(n) to count bits for each number, O(B²) to check pairs where B <= 32 is max bits

⚠ Quadratic Growth

Space Complexity

O(B)

Array to store count of numbers for each bit count, B <= 32

⚡ Linearithmic Space

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Number of Excellent Pairs - Problem

Input & Output

Constraints

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Related Problems

Common Approaches

Bit Count Grouping — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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