Because, there must be t, t<i, sum from 0 to t is sum-S, from t to i is S. Then count of S is count of sum-S. Time Complexity: O(n). n = A.length. Space: O(n). AC Java: 1classSolution {2publicintnumSubarraysWithSum(int[] A,intS) {3if(A ==null|| A.length ...
使用一个字典保存数组某个位置之前的数组和,然后遍历数组求和,这样当我们求到一个位置的和的时候,向前找sum-k是否在数组中,如果在的话,更新结果为之前的结果+1。同时,当前这个sum出现的次数就多了一次。 和560. Subarray Sum Equals K几乎一模一样的题目,为什么就是不会做呢? classSolution(object):defnumSub...
classSolution{public:/** * @param A: an array * @param S: the sum * @return: the number of non-empty subarrays */intnumSubarraysWithSum(vector<int> &A,intS) {// Write your code here.unordered_map<int,int> c({{0,1}});intpsum =0, res =0;for(inti : A) { psum += ...
In an array A of 0s and 1s, how many non-empty subarrays have sum S? Example 1: Input: A = [1,0,1,0,1], S = 2 Output: 4 Explanation: The 4 subarrays are bolded below: [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] [1,0,1,0,1] Note: A.length...
{ public int numSubarraysWithSum(int[] A, int S) { int n = A.length, res = 0, sum = 0; int[] map = new int[n + 1]; map[0] = 1; for (int i = 0; i < n; ++i) { sum += A[i]; if (sum >= S) res += map[sum - S]; ++map[sum]; } return res; } }...
My all DSA question of Leetcode And GFG. Contribute to kunna67/Practice-DSA development by creating an account on GitHub.
so now we need to check whether there is a subarray of a new array ai−λ of length at least x+1 with non-negative sum, which is doable with some prefix sums.Continuous search¶Let f:R→R be a real-valued function that is continuous on a ...
此时还需要考虑一种情况,就是当窗口左边有连续0的时候,因为0并不影响 sum,但是却要算作不同的子数组,所以要统计左起连续0的个数,并且加到结果 res 中即可,参见代码如下: 解法二: classSolution{ public:intnumSubarraysWithSum(vector<int>& A,intS) {intres =0,sum=0, left =0, n = A.size();...
leetcode 930. Binary Subarrays With Sum This remains me of some 'subarray count' type problems….. classSolution{publicintnumSubarraysWithSum(int[] A,intS){int[] ps =newint[A.length +1]; ps[0] =1;intsum=0;intret=0;for(intv: A) {...
publicintnumSubarraysWithSum(int[] A,intS) {intn = A.length, res =0,sum=0;int[]map= newint[n +1];map[0] =1;for(inti =0; i < n; ++i) {sum+= A[i];if(sum>= S) res +=map[sum- S]; ++map[sum]; }returnres; ...