Medium
You are given an array of integers nums
and an integer target
.
Return the number of non-empty subsequences of nums
such that the sum of the minimum and maximum element on it is less or equal to target
. Since the answer may be too large, return it modulo 109 + 7
.
Example 1:
Input: nums = [3,5,6,7], target = 9
Output: 4
Explanation: There are 4 subsequences that satisfy the condition.
[3] -> Min value + max value <= target (3 + 3 <= 9)
[3,5] -> (3 + 5 <= 9)
[3,5,6] -> (3 + 6 <= 9)
[3,6] -> (3 + 6 <= 9)
Example 2:
Input: nums = [3,3,6,8], target = 10
Output: 6
Explanation: There are 6 subsequences that satisfy the condition. (nums can have repeated numbers). [3] , [3] , [3,3], [3,6] , [3,6] , [3,3,6]
Example 3:
Input: nums = [2,3,3,4,6,7], target = 12
Output: 61
Explanation: There are 63 non-empty subsequences, two of them do not satisfy the condition ([6,7], [7]). Number of valid subsequences (63 - 2 = 61).
Constraints:
1 <= nums.length <= 105
1 <= nums[i] <= 106
1 <= target <= 106
class Solution {
fun numSubseq(nums: IntArray, target: Int): Int {
// sorted array will be used to perform binary search
nums.sort()
val mod = 1000000007
// powOf2[i] means (2^i) % mod
val powOf2 = IntArray(nums.size)
powOf2[0] = 1
for (i in 1 until nums.size) {
powOf2[i] = powOf2[i - 1] * 2 % mod
}
var res = 0
var left = 0
var right = nums.size - 1
while (left <= right) {
if (nums[left] + nums[right] > target) {
// nums[right] which is macimum is too big so decrease it
right--
} else {
// every number between right and left be either picked or not picked
// so that is why pow(2, right - left) essentially
res = (res + powOf2[right - left]) % mod
// increment left to find next set of min and max
left++
}
}
return res
}
}