Final Array State After K Multiplication Operations I

Problem Statement

In this problem, you are provided with an integer array nums, an integer k, and another integer multiplier. Your task is to modify the nums array through a series of k operations. Each operation involves finding and manipulating the minimum value in the array. Specifically:

1. Identify the minimum value x in the array nums. If there are multiple occurrences of this minimum, the first instance is selected.
2. The found minimum value x is then multiplied by the multiplier, altering nums at the position of this minimum value.

After completing all k operations, the challenge is to output the altered state of the array nums. This problem is a test of array manipulation, with emphasis on tracking and modifying values based on a set criterion—here being the minimum value finding and its transformation.

Examples

Example 1

Input:

nums = [2,1,3,5,6], k = 5, multiplier = 2

Output:

[8,4,6,5,6]

Explanation:

| Operation | Result |
| --- | --- |
| After operation 1 | [2, 2, 3, 5, 6] |
| After operation 2 | [4, 2, 3, 5, 6] |
| After operation 3 | [4, 4, 3, 5, 6] |
| After operation 4 | [4, 4, 6, 5, 6] |
| After operation 5 | [8, 4, 6, 5, 6] |

Example 2

Input:

nums = [1,2], k = 3, multiplier = 4

Output:

[16,8]

Explanation:

| Operation | Result |
| --- | --- |
| After operation 1 | [4, 2] |
| After operation 2 | [4, 8] |
| After operation 3 | [16, 8] |

Constraints

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100
• 1 <= k <= 10
• 1 <= multiplier <= 5

Approach and Intuition

To devise a solution to the given task, understanding the operation sequence and its effects on the array through iterative steps is essential. Let's break down the process and logic using insights from the provided examples:

• Step 1: Initialize a loop that runs k times to perform the required number of operations.

• Step 2: Within each iteration of the loop, scan through the nums array to find the first occurrence of the current minimum value.

• Step 3: Once the minimum value x is identified, compute x * multiplier and replace the original minimum value with this computed result in the array.

• Step 4: Repeat the process until all k operations are completed. Then, return or print the modified array.

From the examples provided, here’s what we notice:

Observations from Examples

• Observation from Example 1:

  • Input: nums = [2,1,3,5,6], k = 5, multiplier = 2
  • The operations focus on sequentially targeting the new minimum value transformed in each step. Hence, each transformation progressively scales up previously identified minimums.

• Observation from Example 2:

  • Input: nums = [1,2], k = 3, multiplier = 4
  • With fewer elements, the direct repeated scaling of the first element highlights the iterative doubling effect of the operations, with all transformations visible and straightforward.

Given these observations, an algorithm can be framed where the key challenge is the efficient identification of the minimum value and its index, and the in-place update of this value with its scaled version. Moreover, the constraints involving lengths and operation count ensure that the solution remains computationally feasible even with the simplest linear search for the minimum due to small values of k.

Solutions

class Solution {
public:
    vector<int> updateNumbers(vector<int>& data, int iterations, int factor) {
        vector<pair<int, int>> minHeap;
        for (int idx = 0; idx < data.size(); ++idx) {
            minHeap.emplace_back(data[idx], idx);
        }
    
        make_heap(minHeap.begin(), minHeap.end(), greater<>());
    
        while (iterations--) {
            pop_heap(minHeap.begin(), minHeap.end(), greater<>());
            auto [val, pos] = minHeap.back();
            minHeap.pop_back();
    
            data[pos] *= factor;
            minHeap.push_back({data[pos], pos});
            push_heap(minHeap.begin(), minHeap.end(), greater<>());
        }
    
        return data;
    }
};