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LeetCode 930: Binary Subarrays With Sum Solution

Master LeetCode problem 930 (Binary Subarrays With Sum), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

930. Binary Subarrays With Sum

Medium
ArrayHash TableSliding WindowPrefix Sum

Problem Explanation

Explanation:

To solve this problem, we can use the sliding window technique. We iterate through the binary array nums, maintaining a sliding window that keeps track of the sum of elements within the window. At each step, we check if the sum of elements from the start of the window to the current element equals the given goal. If it does, we increment the count of subarrays by 1.

Here are the detailed steps:

  1. Initialize variables count and sum to keep track of the count of subarrays and the running sum respectively.
  2. Initialize a hashmap sumCount to store the frequency of each running sum encountered while iterating through the array.
  3. Iterate through the array, adding each element to the running sum.
  4. For each sum encountered, check if there exists a requiredSum such that sum - requiredSum = goal. If such a requiredSum is found, increment the count by the frequency of requiredSum from the sumCount hashmap.
  5. Increment the frequency of the current sum in the sumCount hashmap.
  6. Return the final count as the answer.

The time complexity of this approach is O(n) where n is the length of the input array nums, and the space complexity is O(n) as well since we use a hashmap to store the frequency of running sums.

:

Solution Code

class Solution {
    public int numSubarraysWithSum(int[] nums, int goal) {
        int count = 0, sum = 0;
        Map<Integer, Integer> sumCount = new HashMap<>();
        sumCount.put(0, 1);
        
        for (int num : nums) {
            sum += num;
            int requiredSum = sum - goal;
            count += sumCount.getOrDefault(requiredSum, 0);
            sumCount.put(sum, sumCount.getOrDefault(sum, 0) + 1);
        }
        
        return count;
    }
}

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