Skip to content

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

Solution 1

Thinking

Count subarrays of a \(0/1\) array whose sum is \(\textit{goal}\). \(n\) is large, so enumerating intervals is too slow. When the prefix sum is \(s\), the number of earlier prefixes equal to \(s-\textit{goal}\) is the number of good subarrays ending here. A counter of prefix frequencies, starting at \(cnt[0]=1\), answers each step in \(O(1)\).

1
2
3
4
5
6
7
8
9
class Solution:
    def numSubarraysWithSum(self, nums: List[int], goal: int) -> int:
        cnt = Counter({0: 1})
        ans = s = 0
        for v in nums:
            s += v
            ans += cnt[s - goal]
            cnt[s] += 1
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
class Solution {
    public int numSubarraysWithSum(int[] nums, int goal) {
        int[] cnt = new int[nums.length + 1];
        cnt[0] = 1;
        int ans = 0, s = 0;
        for (int v : nums) {
            s += v;
            if (s - goal >= 0) {
                ans += cnt[s - goal];
            }
            ++cnt[s];
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution {
public:
    int numSubarraysWithSum(vector<int>& nums, int goal) {
        int cnt[nums.size() + 1];
        memset(cnt, 0, sizeof cnt);
        cnt[0] = 1;
        int ans = 0, s = 0;
        for (int& v : nums) {
            s += v;
            if (s - goal >= 0) {
                ans += cnt[s - goal];
            }
            ++cnt[s];
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
func numSubarraysWithSum(nums []int, goal int) (ans int) {
    cnt := map[int]int{0: 1}
    s := 0
    for _, v := range nums {
        s += v
        ans += cnt[s-goal]
        cnt[s]++
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
/**
 * @param {number[]} nums
 * @param {number} goal
 * @return {number}
 */
var numSubarraysWithSum = function (nums, goal) {
    const cnt = new Array(nums.length + 1).fill(0);
    cnt[0] = 1;
    let ans = 0;
    let s = 0;
    for (const v of nums) {
        s += v;
        if (s >= goal) {
            ans += cnt[s - goal];
        }
        ++cnt[s];
    }
    return ans;
};

Solution 2

Thinking

Method 1 uses linear extra space. Because the array is non-negative, the window sum is monotone in the left end. Two left pointers keep the first index whose sum exceeds \(\textit{goal}\) and the first whose sum is at least \(\textit{goal}\); their difference is the number of windows with sum exactly \(\textit{goal}\).

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
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;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
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;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
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
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
 * @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;
};

Comments