712151. Binary subarray with sum
Binary Subarray with Sum
Summary
Given an array of binary integers, find all subarrays that sum up to a specific target value. The problem involves using a sliding window approach and iterating through the array to identify these subarrays.
Detailed Explanation
To solve this problem, we'll employ a two-pointer technique, also known as the sliding window method. We'll initialize two pointers, start and end, at the beginning of the array. Our goal is to find all subarrays that sum up to the target value.
Here's a step-by-step breakdown:
- Initialize variables:
start: 0 (index of the first element)end: 0 (index of the last element)current_sum: 0 (sum of elements in the current window)target: target sum value
- Iterate through the array:
- While
endis less than the array length:- Add the element at index
endtocurrent_sum - If
current_sumequals thetarget, add the subarray[start, end]to our result list - If
current_sumexceeds thetarget, subtract the element at indexstartfromcurrent_sumand incrementstart
- Add the element at index
- While
- Continue iterating until
endreaches the array length
To illustrate this process, consider an example:
Array: [1, 0, 1, 0, 1, 1]
Target: 2
Result: [[1, 0], [1, 1]]
In this case, we start with start and end at index 0. We add the elements to current_sum, and when it equals the target (2), we add the subarray [1, 0] to our result list. As we move forward, current_sum exceeds the target; so, we subtract the element at index 0 from current_sum and increment start. The process continues until we reach the end of the array.
Optimized Solutions
Java
import java.util.ArrayList;
import java.util.List;
public class BinarySubarray {
public static List<List<Integer>> binarySubarraysWithSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList<>();
int start = 0, end = 0, currentSum = 0;
while (end < nums.length) {
currentSum += nums[end];
if (currentSum == target) {
result.add(List.of(start, end));
currentSum -= nums[start];
start++;
} else if (currentSum > target) {
currentSum -= nums[start];
start++;
}
end++;
}
return result;
}
}