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Minimum Number of Increments on Subarrays to Form a Target Array

Number: 1633

Difficulty: Hard

Paid? No

Companies: Google, Dream11, Uber


Problem Description

Given an integer array target, start with an array initial of the same size filled with zeros. In one operation, you can choose any subarray of initial and increment every element in that subarray by one. The goal is to determine the minimum number of operations required to transform initial into target.


Key Insights

  • You start from an array of zeros and need to increment elements up to the target values.
  • The first element always requires target[0] operations.
  • For subsequent elements, if target[i] is greater than target[i-1], the additional operations required is the difference (target[i] - target[i-1]). When target[i] is less than or equal to target[i-1], no extra operations are needed because the previous operations already overshot and can be adjusted by selecting appropriate subarrays.
  • Main trick: Count increments as the sum of target[0] and all positive differences between consecutive elements.

Space and Time Complexity

Time Complexity: O(n), where n is the number of elements in target. Space Complexity: O(1) (ignoring input storage).


Solution

The solution uses a greedy approach that leverages the observation that the minimum operations needed is equal to the first element of the target array plus the sum of increases between consecutive elements. As you iterate through the array, if the next element is higher than the current one, you add the difference to a running total. No additional operations are necessary when the target value decreases or remains the same since the previous increments already contribute to those indexes.

Data Structures / Algorithms Used:

  • Greedy algorithm
  • Iterative scan of the array

Code Solutions

# Python solution with inline comments
def minNumberOperations(target):
    # Initialize operations with the first element of target (for index 0)
    operations = target[0]
    # Iterate through the target array from the second element onward
    for i in range(1, len(target)):
        # If the current element is greater than the previous element,
        # then we need extra operations equal to the difference.
        if target[i] > target[i - 1]:
            operations += target[i] - target[i - 1]
    return operations

# Example usage:
target = [1, 2, 3, 2, 1]
print(minNumberOperations(target))  # Output: 3
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