# 930. Binary Subarrays With Sum

## Description

Given a binary array nums and an integer goal, return the number of non-empty subarrays with a sum goal.

A subarray is a contiguous part of the array.

Example 1:


Input: nums = [1,0,1,0,1], goal = 2

Output: 4

Explanation: The 4 subarrays are bolded and underlined below:

[1,0,1,0,1]

[1,0,1,0,1]

[1,0,1,0,1]

[1,0,1,0,1]



Example 2:


Input: nums = [0,0,0,0,0], goal = 0

Output: 15



Constraints:

• 1 <= nums.length <= 3 * 104
• nums[i] is either 0 or 1.
• 0 <= goal <= nums.length

## Solutions

• class Solution {
public int numSubarraysWithSum(int[] nums, int goal) {
int i1 = 0, i2 = 0, s1 = 0, s2 = 0, j = 0, ans = 0;
int 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;
}
}

• class Solution {
public:
int numSubarraysWithSum(vector<int>& nums, int goal) {
int i1 = 0, i2 = 0, s1 = 0, s2 = 0, j = 0, ans = 0;
int n = nums.size();
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;
}
};

• 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


• 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;
};