Leetcode 2426 Number of Pairs Satisfying Inequality Solution in Java | Hindi Coding Community

 

You are given two 0-indexed integer arrays nums1 and nums2, each of size n, and an integer diff. Find the number of pairs (i, j) such that:

• 0 <= i < j <= n - 1 and
• nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff.

Return the number of pairs that satisfy the conditions.

 

Example 1:

Input: nums1 = [3,2,5], nums2 = [2,2,1], diff = 1
Output: 3
Explanation:
There are 3 pairs that satisfy the conditions:
1. i = 0, j = 1: 3 - 2 <= 2 - 2 + 1. Since i < j and 1 <= 1, this pair satisfies the conditions.
2. i = 0, j = 2: 3 - 5 <= 2 - 1 + 1. Since i < j and -2 <= 2, this pair satisfies the conditions.
3. i = 1, j = 2: 2 - 5 <= 2 - 1 + 1. Since i < j and -3 <= 2, this pair satisfies the conditions.
Therefore, we return 3.

Example 2:

Input: nums1 = [3,-1], nums2 = [-2,2], diff = -1
Output: 0
Explanation:
Since there does not exist any pair that satisfies the conditions, we return 0.

 

Constraints:

• n == nums1.length == nums2.length
• 2 <= n <= 105
• -104 <= nums1[i], nums2[i] <= 104
• -104 <= diff <= 104

Java Code :

public long numberOfPairs(int[] nums1, int[] nums2, int diff) {

int n = nums1.length;

long[] nums = new long[n];

for (int i = 0; i < n; i++) nums[i] = nums1[i] - nums2[i];

long[] tmp = Arrays.copyOf(nums, n);

return sort(nums, tmp, 0, n-1, diff);

}

// sort O(NlogN)

private long sort(long[] nums, long[] tmp, int lo, int hi, int diff) {

if (lo == hi) return 0L;

int mid = lo + (hi - lo) / 2;

long res = 0L;

res += sort(nums, tmp, lo, mid, diff);

res += sort(nums, tmp, mid+1, hi, diff);

res += merge(nums, tmp, lo, mid, hi, diff);

return res;

}

// merge 2*N = O(N)

private long merge(long[] nums, long[] tmp, int lo, int mid, int hi, int diff) {

long res = 0L;

// find the count of pairs x <= y + diff in 2 sorted halves

for (int i = lo, j = mid+1; i <= mid && j <= hi; i++) {

while (j <= hi && nums[i] > nums[j] + diff) j++;

res += hi-j+1;

}

// merge 2 sorted halves

for (int i = lo, j = mid+1, k = lo; i <= mid || j <= hi; k++) {

if (j > hi || (i <= mid && nums[i] < nums[j])) {

tmp[k] = nums[i++];

} else {

tmp[k] = nums[j++];

}

}

System.arraycopy(tmp, lo, nums, lo, hi-lo+1); // tmp -> nums

return res;

}