Given a 0-indexed integer array nums of size n and two integers lower and upper, return the number of fair pairs.
A pair (i, j) is fair if:
0 <= i < j < n, andlower <= nums[i] + nums[j] <= upper
Example 1:
Input: nums = [0,1,7,4,4,5], lower = 3, upper = 6 Output: 6 Explanation: There are 6 fair pairs: (0,3), (0,4), (0,5), (1,3), (1,4), and (1,5).
Example 2:
Input: nums = [1,7,9,2,5], lower = 11, upper = 11 Output: 1 Explanation: There is a single fair pair: (2,3).
Constraints:
1 <= nums.length <= 105nums.length == n-109 <= nums[i] <= 109-109 <= lower <= upper <= 109class Solution {
public long countFairPairs(int[] nums, int lower, int upper) {
Arrays.sort(nums);
return lower_bound(nums, upper + 1) - lower_bound(nums, lower);
}
// Calculate the number of pairs with sum less than `value`.
private long lower_bound(int[] nums, int value) {
int left = 0, right = nums.length - 1;
long result = 0;
while (left < right) {
int sum = nums[left] + nums[right];
// If sum is less than value, add the size of window to result and move to the
// next index.
if (sum < value) {
result += (right - left);
left++;
} else {
// Otherwise, shift the right pointer backwards, until we get a valid window.
right--;
}
}
return result;
}
}class Solution {
public long countFairPairs(int[] nums, int lower, int upper) {
Arrays.sort(nums);
return lower_bound(nums, upper + 1) - lower_bound(nums, lower);
}
// Calculate the number of pairs with sum less than `value`.
private long lower_bound(int[] nums, int value) {
int left = 0, right = nums.length - 1;
long result = 0;
while (left < right) {
int sum = nums[left] + nums[right];
// If sum is less than value, add the size of window to result and move to the
// next index.
if (sum < value) {
result += (right - left);
left++;
} else {
// Otherwise, shift the right pointer backwards, until we get a valid window.
right--;
}
}
return result;
}
}