DevExCode

2563. Count the Number of Fair Pairs

ArrayTwo PointersBinary SearchSorting

Explanation:

To solve this problem, we can iterate through all pairs of indices (i, j) with 0 <= i < j < n and count the pairs that satisfy the condition lower <= nums[i] + nums[j] <= upper. We can achieve this by maintaining a HashMap to store the count of each prefix sum from the beginning of the array up to the current index.

  1. Initialize a HashMap prefixSums to store the count of each prefix sum.
  2. Initialize a variable fairPairs to keep track of the number of fair pairs.
  3. Iterate through the array nums and for each element at index j:
    • Calculate the current prefix sum currentSum by adding nums[j] to the previous prefix sum.
    • Increment the fairPairs count by the value in prefixSums.getOrDefault(currentSum - lower, 0).
    • Increment the count of currentSum in the prefixSums HashMap.
  4. Return the total count of fair pairs.

Time Complexity: O(n), where n is the length of the input array nums. Space Complexity: O(n) for the HashMap prefixSums.

:

View Leetcode Problem
class Solution {
    public int countPairs(int[] nums, int lower, int upper) {
        Map<Integer, Integer> prefixSums = new HashMap<>();
        int fairPairs = 0;
        for (int j = 0; j < nums.length; j++) {
            int currentSum = nums[j] + (j > 0 ? nums[j - 1] : 0);
            for (int i = 0; i < j; i++) {
                if (lower <= currentSum && currentSum <= upper) {
                    fairPairs++;
                }
                currentSum += nums[i];
            }
            prefixSums.put(currentSum, prefixSums.getOrDefault(currentSum, 0) + 1);
        }
        return fairPairs;
    }
}

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