LeetCode in Kotlin
2824. Count Pairs Whose Sum is Less than Target
Easy
Given a 0-indexed integer array nums of length n and an integer target, return the number of pairs (i, j) where 0 <= i < j < n and nums[i] + nums[j] < target.
Example 1:
Input: nums = [-1,1,2,3,1], target = 2
Output: 3
Explanation:
There are 3 pairs of indices that satisfy the conditions in the statement:
-
(0, 1) since 0 < 1 and nums[0] + nums[1] = 0 < target
-
(0, 2) since 0 < 2 and nums[0] + nums[2] = 1 < target
-
(0, 4) since 0 < 4 and nums[0] + nums[4] = 0 < target
Note that (0, 3) is not counted since nums[0] + nums[3] is not strictly less than the target.
Example 2:
Input: nums = [-6,2,5,-2,-7,-1,3], target = -2
Output: 10
Explanation:
There are 10 pairs of indices that satisfy the conditions in the statement:
-
(0, 1) since 0 < 1 and nums[0] + nums[1] = -4 < target
-
(0, 3) since 0 < 3 and nums[0] + nums[3] = -8 < target
-
(0, 4) since 0 < 4 and nums[0] + nums[4] = -13 < target
-
(0, 5) since 0 < 5 and nums[0] + nums[5] = -7 < target
-
(0, 6) since 0 < 6 and nums[0] + nums[6] = -3 < target
-
(1, 4) since 1 < 4 and nums[1] + nums[4] = -5 < target
-
(3, 4) since 3 < 4 and nums[3] + nums[4] = -9 < target
-
(3, 5) since 3 < 5 and nums[3] + nums[5] = -3 < target
-
(4, 5) since 4 < 5 and nums[4] + nums[5] = -8 < target
-
(4, 6) since 4 < 6 and nums[4] + nums[6] = -4 < target
Constraints:
1 <= nums.length == n <= 50-50 <= nums[i], target <= 50
Solution
class Solution {
fun countPairs(nums: List<Int>, target: Int): Int {
var cnt = 0
for (i in nums.indices) {
for (j in i + 1 until nums.size) {
if (nums[i] + nums[j] < target) {
cnt++
}
}
}
return cnt
}
}