Minimum Total Operations solution and explanation
The problem
Given an array of integers nums, you can perform any number of operations on this array.
In each operation, you can:
- Choose a prefix of the array.
- Choose an integer
k(which can be negative) and addkto each element in the chosen prefix.
A prefix of an array is a subarray that starts from the beginning of the array and extends to any point within it.
Return the minimum number of operations required to make all elements in arr equal.
Example 1:
Input: nums = [1,4,2]
Output: 2
Explanation:
- Operation 1: Choose the prefix
[1, 4]of length 2 and add -2 to each element of the prefix. The array becomes[-1, 2, 2]. - Operation 2: Choose the prefix
[-1]of length 1 and add 3 to it. The array becomes[2, 2, 2]. - Thus, the minimum number of required operations is 2.
Example 2:
Input: nums = [10,10,10]
Output: 0
Explanation:
- All elements are already equal, so no operations are needed.
Constraints:
1 <= nums.length <= 105-109 <= nums[i] <= 109
Approach overview
Problem Overview: You are given an integer array and need to compute the minimum number of operations required to make the array satisfy a required ordering constraint. Each operation modifies an element, and the goal is to minimize the total operations across the entire array.
Approach 1: Single Pass Greedy Scan (O(n) time, O(1) space)
The most efficient strategy is a greedy single pass through the array. Traverse from left to right while tracking the previous valid value. If the current value already satisfies the required condition relative to the previous element, move forward. If it violates the condition, compute how many operations are needed to adjust it and add that to the running total. After adjustment, treat the corrected value as the new reference for the next step. This works because fixing violations locally during a single traversal guarantees the global minimum number of operations.
The key insight is that you never need to revisit earlier elements. Each element only depends on the constraint with its immediate predecessor, so a forward scan is sufficient. That eliminates the need for nested loops or backtracking. Operations accumulate based on the difference between the current value and the minimum value required to maintain the constraint.
This pattern appears frequently in array problems where you enforce ordering or monotonic properties. Greedy adjustments during iteration keep the algorithm linear while maintaining correctness.
Recommended for interviews: Interviewers expect the single pass greedy solution. A brute force approach that repeatedly scans or adjusts the array demonstrates understanding of the problem constraints, but the optimal solution shows you recognize that each element only depends on the previous state. Achieving O(n) time with O(1) extra space is the key optimization for most array traversal problems and demonstrates strong algorithmic reasoning.
Solution
We can traverse the array, and for each element, if it is not equal to the previous element, we need to perform an operation. Finally, we return the number of operations.
The time complexity is O(n), where n is the length of the array. The space complexity is O(1).
class Solution:
def minOperations(self, nums: List[int]) -> int:
return sum(x != y for x, y in pairwise(nums))class Solution {
public int minOperations(int[] nums) {
int ans = 0;
for (int i = 1; i < nums.length; ++i) {
ans += nums[i] != nums[i - 1] ? 1 : 0;
}
return ans;
}
}class Solution {
public:
int minOperations(vector<int>& nums) {
int ans = 0;
for (int i = 1; i < nums.size(); ++i) {
ans += nums[i] != nums[i - 1];
}
return ans;
}
};/**
* @param {number[]} nums
* @return {number}
*/
var minOperations = function (nums) {
let ans = 0;
for (let i = 1; i < nums.length; ++i) {
ans += nums[i] !== nums[i - 1] ? 1 : 0;
}
return ans;
};function minOperations(nums: number[]): number {
let ans = 0;
for (let i = 1; i < nums.length; ++i) {
ans += nums[i] !== nums[i - 1] ? 1 : 0;
}
return ans;
}func minOperations(nums []int) (ans int) {
for i, x := range nums[1:] {
if x != nums[i] {
ans++
}
}
return
}impl Solution {
pub fn min_operations(nums: Vec<i32>) -> i32 {
let mut ans = 0;
for i in 1..nums.len() {
if nums[i] != nums[i - 1] {
ans += 1;
}
}
ans
}
}Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Repeated Adjustment / Brute Force | O(n^2) | O(1) | Useful for understanding the constraint by repeatedly fixing violations in the array. |
| Single Pass Greedy Scan | O(n) | O(1) | Best approach for interviews and production. One linear traversal computes minimal operations. |
Frequently asked questions
Is Minimum Total Operations easy or hard?
Minimum Total Operations is generally considered an easy problem because it relies on a straightforward greedy observation and a single linear scan. Once you recognize that each element only depends on the previous one, the implementation becomes simple and efficient.
Minimum Total Operations Python/Java solution
The Python and Java implementations both follow the same logic: iterate through the array once, track the previous valid value, and accumulate operations whenever the current value needs adjustment. Because the algorithm is language-independent, the time complexity remains O(n) with O(1) space.
How to solve Minimum Total Operations in O(n)?
Traverse the array once while maintaining the previous valid value. If the current element satisfies the constraint, continue. If it violates the rule, compute the difference required to fix it, add that difference to the total operations, and update the current value accordingly. This single-pass greedy process ensures O(n) time complexity.
What is the best approach for Minimum Total Operations?
The optimal approach is a greedy single-pass traversal of the array. Iterate from left to right and adjust elements whenever they violate the required ordering constraint with the previous element. This guarantees the minimal number of operations while running in O(n) time and O(1) extra space.
Is Minimum Total Operations asked at Google/Amazon/Meta?
Problems that require minimizing operations on arrays using greedy traversal patterns frequently appear in interviews at companies like Amazon, Google, and Meta. Variations of this pattern test whether candidates can recognize when a single-pass adjustment strategy replaces expensive repeated scans.
What data structure is used in Minimum Total Operations?
The primary data structure is a simple array. The algorithm processes it sequentially using a greedy strategy and a few tracking variables. No additional structures like heaps or hash maps are required, which keeps the space complexity constant.
What is the time complexity of Minimum Total Operations?
The optimal solution runs in O(n) time because the array is processed exactly once. Each element is checked against its predecessor and adjusted if needed. The space complexity is O(1) since only a few variables are used for tracking the previous value and operation count.