LeetCode 2578: Split With Minimum Sum Solution
Master LeetCode problem 2578 (Split With Minimum Sum), a easy 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.
Problem Explanation
Explanation
To solve this problem, we can follow these steps:
- Iterate through the digits of the given number from left to right.
- For each digit, assign it to one of the two numbers such that the sum of each digit's occurrence in the two numbers equals the total occurrence in the given number.
- Keep track of the minimum sum obtained by comparing the current sum after each assignment.
The time complexity of this approach is O(N), where N is the number of digits in the given number. The space complexity is O(1) as we are using a constant amount of extra space.
Solution Code
class Solution {
public int minSumOfLengths(int[] arr, int target) {
Map<Integer, Integer> map = new HashMap<>();
map.put(0, -1);
int sum = 0, minLen = Integer.MAX_VALUE, result = Integer.MAX_VALUE;
Map<Integer, Integer> dp = new HashMap<>();
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
map.put(sum, i);
}
sum = 0;
for (int i = 0; i < arr.length; i++) {
sum += arr[i];
if (map.containsKey(sum - target)) {
int len1 = i - map.get(sum - target);
int len2 = dp.getOrDefault(map.get(sum - target), Integer.MAX_VALUE);
if (len2 != Integer.MAX_VALUE) {
result = Math.min(result, len1 + len2);
}
minLen = Math.min(minLen, len1);
dp.put(i, minLen);
}
}
return result == Integer.MAX_VALUE ? -1 : result;
}
}Try It Yourself
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