LeetCode 2919: Minimum Increment Operations to Make Array Beautiful Solution
Master LeetCode problem 2919 (Minimum Increment Operations to Make Array Beautiful), 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.
2919. Minimum Increment Operations to Make Array Beautiful
Problem Explanation
Explanation
To make the array beautiful, we need to ensure that for any subarray of size 3 or more, the maximum element is greater than or equal to k. To achieve this, we can iterate over the array and for each element, we calculate the required increment to make it greater than or equal to k while maintaining the minimum overall increment operations required. We can do this by keeping track of the current maximum element in the subarray so far.
Algorithm
- Initialize
opsto 0 to keep track of the total increment operations needed. - Initialize
maxSeento 0 to keep track of the current maximum element in the subarray so far. - Iterate over the array:
- If the current element is less than
k, calculate the increment required to make it greater than or equal tok. - If the current element is greater than or equal to
k, updatemaxSeen. - Increment
opsby the difference betweenmaxSeenand the current element.
- If the current element is less than
- Return
ops.
Time Complexity
The time complexity of this algorithm is O(n), where n is the length of the input array.
Space Complexity
The space complexity of this algorithm is O(1) as we are using only a constant amount of extra space.
Solution Code
class Solution {
public int minOperations(int[] nums, int k) {
int ops = 0;
int maxSeen = 0;
for (int num : nums) {
if (num < k) {
ops += k - num;
} else {
maxSeen = Math.max(maxSeen, num);
}
ops += Math.max(0, maxSeen - num);
}
return ops;
}
}Try It Yourself
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