883232. Minimum Removals for Target Sum
Minimum Removals for Target Sum
Summary
The problem "Minimum Removals for Target Sum" involves given an array and a target sum, find the minimum number of elements that need to be removed from the array so that the remaining elements sum up to the target. This is a classic example of a problem that can be solved using the sliding window technique.
Detailed Explanation
To solve this problem, we will use a two-pointer approach. We initialize two pointers, left and right, both pointing to the start of the array. The idea is to maintain a sum of elements within the current window [left, right] that equals the target sum.
Here's the step-by-step breakdown:
- Initialize
leftandrightto 0. - Calculate the sum of the first
right - left + 1elements (arr[left:right+1]) and store it in a variablecurrentSum. - If
currentSumequals the target sum, returnright - left + 1(the number of elements in the current window). - If
currentSumis less than the target sum, incrementrightto move the window forward and recalculatecurrentSum. - If
currentSumis greater than the target sum, decrementleftto shrink the window from the left side and recalculatecurrentSum. - Repeat steps 2-5 until
currentSumequals the target sum or we reach the end of the array. - Return the minimum number of elements that need to be removed (which is the difference between the length of the original array and the length of the window where the sum equals the target).
Time complexity: O(n), where n is the length of the input array.
Space complexity: O(1), as we only use a constant amount of space to store the pointers and the sum.
Optimized Solutions
Java
public int minRemovals(int[] arr, int targetSum) {
int left = 0;
int right = 0;
int currentSum = 0;
int minRemovals = Integer.MAX_VALUE;
while (right < arr.length) {
currentSum += arr[right];
while (currentSum > targetSum && left <= right) {
currentSum -= arr[left++];
}
if (currentSum == targetSum) {
minRemovals = Math.min(minRemovals, right - left + 1);
}
right++;
}
return minRemovals;
}