LeetCode 1827: Minimum Operations to Make the Array Increasing

Explanation

To solve this problem, we need to iterate through the array and for each element, we check if it is less than or equal to the previous element. If it is, we increment the current element to make it strictly increasing. The number of operations needed is the difference between the current element and the previous element plus 1. We keep track of the total operations needed and update the previous element for the next iteration.

Algorithm:

1. Initialize a variable prev to store the previous element.
2. Initialize a variable operations to store the total operations needed.
3. Iterate through the array starting from index 1.
4. For each element, if it is less than or equal to the previous element, increment it to make it strictly increasing. Add the difference between the updated element and the previous element to operations.
5. Update prev to the updated element.
6. Return the total operations needed.

Time Complexity: The time complexity of this algorithm is O(n) where n is the number of elements in the input array.

Space Complexity: The space complexity is O(1) as we are using only a constant amount of extra space.

class Solution {
    public int minOperations(int[] nums) {
        int operations = 0;
        int prev = nums[0];
        
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] <= prev) {
                operations += prev - nums[i] + 1;
                prev++;
            } else {
                prev = nums[i];
            }
        }
        
        return operations;
    }
}