Question

930. Binary Subarrays With Sum

中文文档

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

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

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;
  }
}

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;
  }
};

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
}

/**
 * @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

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 {
  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;
  }
};

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;
};