This "Easy" level problem asks for the minimum operations to make an array strictly increasing. In one operation, you can increment any element of the array by 1. Since you want to use the minimum total operations, for each element, you should only increment it to the smallest possible value that is strictly greater than the element before it.
Companies like Deutsche Bank and Amazon use this to test basic array traversal and the Greedy property. It's a foundational problem that ensures a candidate can correctly implement a one-pass algorithm and maintain a running requirement (the strictly increasing constraint) without unnecessary complexity.
The primary pattern is the Greedy approach. As you traverse the array from left to right starting from the second element, compare the current element arr[i] with the previous element arr[i-1]. If arr[i] is not greater than arr[i-1], you must increase arr[i] to arr[i-1] + 1. The number of operations for this step is (arr[i-1] + 1) - arr[i]. Update the element and add to the total count.
Array: [1, 5, 2, 4, 1].
6 - 2 = 4. Array: [1, 5, 6, 4, 1].7 - 4 = 3. Array: [1, 5, 6, 7, 1].8 - 1 = 7. Array: [1, 5, 6, 7, 8].
Total operations = 4 + 3 + 7 = 14.A common error is trying to find the maximum element first and making everything greater than that, which uses more operations than necessary. Another mistake is failing to update the "previous" value in the loop, leading to incorrect comparisons. Some candidates also use a second array instead of modifying the existing one or just keeping track of the "last seen" required value, which adds unnecessary space complexity.
Always remember the greedy principle for "minimum operations to make non-decreasing/increasing" problems: the best way to satisfy the condition A < B is to make B as small as possible, which is A + 1. This logic ensures that B puts the least possible constraint on the elements that follow it.