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Solution.java
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Solution.java
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package g1401_1500.s1498_number_of_subsequences_that_satisfy_the_given_sum_condition;
// #Medium #Array #Sorting #Binary_Search #Two_Pointers #Binary_Search_II_Day_15
// #2022_03_24_Time_27_ms_(99.13%)_Space_50.5_MB_(97.89%)
import java.util.Arrays;
public class Solution {
public int numSubseq(int[] nums, int target) {
// sorted array will be used to perform binary search
Arrays.sort(nums);
int mod = 1000_000_007;
// powOf2[i] means (2^i) % mod
int[] powOf2 = new int[nums.length];
powOf2[0] = 1;
for (int i = 1; i < nums.length; i++) {
powOf2[i] = (powOf2[i - 1] * 2) % mod;
}
int res = 0;
int left = 0;
int right = nums.length - 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;
}
}