Minimum Total Operations - Practice Coding Problems

Minimum Total Operations - Problem

Array Easy

Minimum Total Operations

Imagine you have an array of integers and a powerful tool that lets you modify prefixes of the array. In each operation, you can:

1. Choose any prefix of the array (a subarray starting from index 0)
2. Choose any integer k (positive, negative, or zero)
3. Add k to every element in that prefix

Your goal is to make all elements in the array equal using the minimum number of operations.

Example: Given [3, 1, 4], you could:
• Operation 1: Add -2 to prefix [3, 1] → [1, -1, 4]
• Operation 2: Add 2 to prefix [-1] → [1, 1, 4]
• Operation 3: Add -3 to prefix [4] → [1, 1, 1]

Find the optimal strategy to equalize all elements!

Input & Output

example_1.py — Basic Case

$ Input: [2, 3, 3, 1, 1, 1]

› Output: 3

💡 Note: We need 3 operations because there are 3 positions where adjacent elements differ: 2≠3 (pos 0-1), 3≠1 (pos 2-3), and these differences require separate operations to resolve.

example_2.py — All Equal

$ Input: [1, 1, 1, 1]

› Output: 0

💡 Note: All elements are already equal, so no operations are needed. Every adjacent pair is identical.

example_3.py — All Different

$ Input: [5, 4, 3, 2, 1]

› Output: 4

💡 Note: Every adjacent pair is different (5≠4, 4≠3, 3≠2, 2≠1), so we need 4 operations to make all elements equal to the rightmost value.

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁹ ≤ nums[i] ≤ 10⁹
At most 10⁵ operations will be needed

Visualization

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Understanding the Visualization

Identify the Pattern

Operations affect prefixes, like paint flowing down stairs

Choose Target Wisely

The rightmost element naturally becomes our target

Count Differences

Each adjacent difference needs one operation to fix

Apply the Formula

Total operations = number of adjacent differences

Key Takeaway

🎯 Key Insight: The minimum operations equals the number of adjacent differences because each difference requires exactly one prefix operation to resolve!

Asked in

G Google 28 a Amazon 22 f Meta 19 ⊞ Microsoft 15

The optimal solution works by scanning the array from right to left and counting adjacent differences. Since operations affect prefixes, each difference between adjacent elements requires exactly one operation to resolve. This gives us O(n) time and O(1) space complexity.

Common Approaches

Approach	Time	Space	Notes
✓ Optimal Solution (Right to Left)	O(n)	O(1)	Work from right to left, count operations when adjacent elements differ
Brute Force (Try All Targets)	O(n²)	O(n)	Try each possible target value and calculate minimum operations

Optimal Solution (Right to Left) — Algorithm Steps

Start from the rightmost element (this will be our final target)
Move left, comparing each element with the next one
When elements differ, we need one operation to make them equal
Count total operations needed
Return the count

Visualization

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Step-by-Step Walkthrough

Start from Right

Begin at rightmost element as target

Compare Adjacent

Check if current differs from next element

Count Operation

If different, increment operation counter

Move Left

Continue to previous element

Return Count

Total operations equals total differences

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int minimumOperations(int* nums, int n) {
    int operations = 0;
    
    // Work from right to left
    for (int i = n - 2; i >= 0; i--) {
        // If current element differs from next element,
        // we need an operation
        if (nums[i] != nums[i + 1]) {
            operations++;
        }
    }
    
    return operations;
}

int main() {
    int nums[1000];
    int n = 0;
    char input[10000];
    fgets(input, sizeof(input), stdin);
    
    char* token = strtok(input, "[,]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", minimumOperations(nums, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the array from right to left

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable, no extra data structures

✓ Linear Space

27.5K Views

Medium Frequency

~15 min Avg. Time

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimal Solution (Right to Left) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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