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Minimum Total Operations

Number: 3694

Difficulty: Easy

Paid? Yes

Companies: N/A

Problem Description

Given an integer array nums, you are allowed to perform any number of operations. In each operation, you can choose a prefix (a subarray starting from the beginning) and add any integer k (which can be negative) to every element in that prefix. The goal is to determine the minimum number of operations needed to make every element in the array equal.

Key Insights

  • The last element of the array is never affected by any operation on a proper prefix, so it makes sense to convert the entire array to the last element.
  • Working backwards, if an element is different from its next neighbor, then an operation is needed to change that prefix so that the current element becomes equal to the next one.
  • The minimal number of operations is therefore equal to the number of indices i (from 0 to n-2) where nums[i] differs from nums[i+1].

Space and Time Complexity

Time Complexity: O(n), where n is the number of elements in nums. Space Complexity: O(1), as only a constant amount of extra space is used.

Solution

The strategy is to make all elements equal to the last element in the array because prefix operations will not affect the last element. Iterate from the beginning of the array to the second-to-last element. For each adjacent pair (nums[i] and nums[i+1]), if they are not equal, an operation is required to fix the discrepancy. Therefore, the answer is simply the count of distinct transitions between consecutive elements.

Data structures are minimal (just a loop counter and index iteration), and the algorithm is a direct linear scan using a for loop. This approach ensures optimal time efficiency and is straightforward based on the observation that each difference signals the need for one operation to “correct” that block of the array.

Code Solutions

# Python solution for Minimum Total Operations
def minOperations(nums):
    # Initialize the operation count to 0
    operations = 0
    # Iterate through the array from index 0 to n-2
    for i in range(len(nums) - 1):
        # If the current element does not equal the next element,
        # an operation is required
        if nums[i] != nums[i + 1]:
            operations += 1
    return operations

# Example usage:
if __name__ == "__main__":
    print(minOperations([1, 4, 2]))  # Expected output: 2
    print(minOperations([10, 10, 10]))  # Expected output: 0
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