LeetCode 2163: Minimum Difference in Sums After Removal of Elements Solution
Master LeetCode problem 2163 (Minimum Difference in Sums After Removal of Elements), a hard 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.
2163. Minimum Difference in Sums After Removal of Elements
Problem Explanation
Explanation
To solve this problem, we need to find the minimum difference between the sums of two parts after removing n elements. The key idea is to iterate through all possible combinations of removing n elements, calculating the sum of each part, and keeping track of the minimum difference found so far.
We can achieve this by using a recursive function that simulates removing elements from the array. At each step, we either choose to remove the current element or skip it. We continue this process until we have removed n elements, at which point we calculate the sums of the two parts and update the minimum difference.
The time complexity of this approach is O(2^n) because we are exploring all possible combinations of removing elements. The space complexity is O(n) for the recursive call stack.
Solution Code
class Solution {
public int minimumDifference(int[] nums) {
int n = nums.length / 3;
return dfs(nums, 0, 0, 0, n);
}
private int dfs(int[] nums, int index, int selected, int sum, int n) {
if (index == nums.length) {
if (selected == n) {
return Math.abs(sum);
}
return Integer.MAX_VALUE;
}
int choose = dfs(nums, index + 1, selected + 1, sum + nums[index], n);
int skip = dfs(nums, index + 1, selected, sum - nums[index], n);
return Math.min(choose, skip);
}
}Try It Yourself
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