Minimum Number of Increments on Subarrays to Form a Target Array - Solution & Explanation

To solve the problem optimally, we can use the concept of differences between consecutive elements in the target array. The number of operations required to achieve the target at each index is dictated by the difference between the current element and the previous element in the target array. By summing all the positive differences, we get the total number of operations needed.

Here's why this works: starting from the initial array (which is all zeros), each increment can be seen as targeting a 'range' of the array. The increments impose certain requirements for transitions between consecutive numbers. Therefore, the sum of necessary transitions gives us the minimum operations required.

The solution starts by identifying the minimum number of operations needed as the first element of the target since we have to create that amount starting from zero. As we iterate over the target array, if a position 'i' has a value larger than its previous position, it's necessary to increment the difference to account for it. The sum of those required transitions is our answer.

Another approach to solve this problem is using a greedy strategy that incrementally builds the target array value-by-value, choosing increments as needed to transition between values in the target. This method efficiently considers the necessary increments at each step by analyzing both the current and previous elements.

This is done by incrementally forming the target using the fewest number of operations required when seen as a series of number levels that need reaching.

In this C solution, the function iterates through the target array while maintaining a 'previous' value that starts at zero. For each target element, if it's greater than 'previous', the difference is added to operations, effectively counting increments needed.