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 <= 1051 <= nums[i] <= 1061 <= target <= 106class Solution {
int MOD = 1_000_000_007;
public int numSubseq(int[] nums, int target) {
Arrays.sort(nums); // order doesnt matter
int n = nums.length, end = n - 1, ans = 0, k = 1;
// precalculate modded powers of 2
int[] powersOf2 = new int[n];
powersOf2[0] = 1;
for (int i = 0; i < n; i++) {
int bs = binarySearch(nums, i, end, target);
if (bs < 0)
break;
// System.out.println(i + "\t" + bs);
end = bs; // decrease search space
for (; k <= bs; k++)
powersOf2[k] = (powersOf2[k - 1] << 1) % MOD;
ans = (ans + powersOf2[bs - i]) % MOD;
}
return ans;
}
// find valid key in bounded search space
private int binarySearch(int[] nums, int lt, int rt, int target) {
int key = target - nums[lt], bs = -1;
while (lt <= rt) {
int mid = (lt + rt) / 2;
if (nums[mid] <= key) {
bs = mid;
lt = mid + 1;
} else
rt = mid - 1;
}
return bs;
}
}class Solution {
int MOD = 1_000_000_007;
public int numSubseq(int[] nums, int target) {
Arrays.sort(nums); // order doesnt matter
int n = nums.length, end = n - 1, ans = 0, lt = 0, rt = n - 1;
// precalculate modded powers of 2
int[] powersOf2 = new int[n];
powersOf2[0] = 1;
for (int i = 1; i < n; i++)
powersOf2[i] = (powersOf2[i - 1] << 1) % MOD;
// binary search
while (lt <= rt)
if (nums[lt] + nums[rt] <= target)
ans = (ans + powersOf2[rt - lt++]) % MOD;
else
rt--;
return ans;
}
}