Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/930.html

930. Binary Subarrays With Sum

Level

Medium

Description

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:

  1. A.length <= 30000
  2. 0 <= S <= A.length
  3. A[i] is either 0 or 1.

Solution

Use a map to count the number of subarrays starting from index 0 that have sum equal to a given value.

If S is 0, then for each possible sum in the map, count the number of subarrays of the sum. If there are k such subarrays, then the number of non-empty subarrays obtained from such subarrays is k * (k - 1) / 2. Count the number of non-empty subarrays with sum 0 from all possible sums in the map.

If S is positive, then for each pair (i, i + S) such that i >= 0 and i + S doesn’t exceed the maximum possible sum (which is the sum of all numbers in the array), let count1 and count2 be the numbers of subarrays starting from index 0 that have sum i and sum i + S respectively, and the number of non-empty subarrays with sum S for the pair (i, i + S) is count1 * count2. Count the number of non-empty subarrays with sum S from all possible sums in the map.

  • class Solution {
    public int numSubarraysWithSum(int[] A, int S) {
        if (A == null || A.length == 0)
            return 0;
        Map<Integer, Integer> sumCountMap = new HashMap<Integer, Integer>();
        int length = A.length;
        int beginIndex = 0;
        int curSum = A[0];
        for (int i = 1; i < length; i++) {
            if (A[i] == 1) {
                sumCountMap.put(curSum, i - beginIndex);
                curSum++;
                beginIndex = i;
            }
        }
        sumCountMap.put(curSum, length - beginIndex);
        int totalCount = 0;
        sumCountMap.put(0, sumCountMap.getOrDefault(0, 0) + 1);
        if (S == 0) {
            Set<Integer> keySet = sumCountMap.keySet();
            for (int num : keySet) {
                int count = sumCountMap.get(num);
                totalCount += count * (count - 1) / 2;
            }
            return totalCount;
        } else {
            int minSum = A[0], maxSum = curSum;
            int upper = maxSum - S;
            for (int i = 0; i <= upper; i++) {
                int count1 = sumCountMap.get(i), count2 = sumCountMap.get(i + S);
                totalCount += count1 * count2;
            }
            return totalCount;
        }
    }
}

  • // OJ: https://leetcode.com/problems/binary-subarrays-with-sum/
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int numSubarraysWithSum(vector<int>& A, int goal) {
        unordered_map<int, int> m{ {0,-1} }; // count of 1s -> index
        int ans = 0;
        for (int i = 0, N = A.size(), sum = 0; i < N; ++i) {
            sum += A[i];
            if (m.count(sum) == 0) m[sum] = i;
            if (goal == 0) ans += i - m[sum];
            else if (sum - goal >= 0) ans += m[sum - goal + 1] - m[sum - goal];
        }
        return ans;
    }
};

  • class Solution:
    def numSubarraysWithSum(self, nums: List[int], goal: int) -> int:
        i1 = i2 = s1 = s2 = j = ans = 0
        n = len(nums)
        while j < n:
            s1 += nums[j]
            s2 += nums[j]
            while i1 <= j and s1 > goal:
                s1 -= nums[i1]
                i1 += 1
            while i2 <= j and s2 >= goal:
                s2 -= nums[i2]
                i2 += 1
            ans += i2 - i1
            j += 1
        return ans

############

class Solution(object):
    def numSubarraysWithSum(self, A, S):
        """
        :type A: List[int]
        :type S: int
        :rtype: int
        """
        N = len(A)
        tsum = [0] * (N + 1)
        for i in range(1, N + 1):
            tsum[i] = tsum[i - 1] + A[i - 1]
        res = 0
        for i in range(1, N + 1):
            remain = tsum[i] - S
            if remain < 0:
                continue
            left = bisect.bisect_left(tsum, remain)
            right = bisect.bisect_right(tsum, remain)
            right = min(i, right)
            res += right - left
        return res

  • func numSubarraysWithSum(nums []int, goal int) int {
	i1, i2, s1, s2, j, ans, n := 0, 0, 0, 0, 0, 0, len(nums)
	for j < n {
		s1 += nums[j]
		s2 += nums[j]
		for i1 <= j && s1 > goal {
			s1 -= nums[i1]
			i1++
		}
		for i2 <= j && s2 >= goal {
			s2 -= nums[i2]
			i2++
		}
		ans += i2 - i1
		j++
	}
	return ans
}

  • /**
 * @param {number[]} nums
 * @param {number} goal
 * @return {number}
 */
var numSubarraysWithSum = function (nums, goal) {
    let i1 = 0,
        i2 = 0,
        s1 = 0,
        s2 = 0,
        j = 0,
        ans = 0;
    const n = nums.length;
    while (j < n) {
        s1 += nums[j];
        s2 += nums[j];
        while (i1 <= j && s1 > goal) s1 -= nums[i1++];
        while (i2 <= j && s2 >= goal) s2 -= nums[i2++];
        ans += i2 - i1;
        ++j;
    }
    return ans;
};

All Problems

All Solutions