3091. Apply Operations to Make Sum of Array Greater Than or Equal to k

MediumGreedyMathEnumeration

Problem Description

You are given a positive integer k. Initially, you have an array nums = [1]. You can perform any of the following operations on the array any number of times (possibly zero):

Choose any element in the array and increase its value by 1.
Duplicate any element in the array and add it to the end of the array.

Return the minimum number of operations required to make the sum of elements of the final array greater than or equal to k.

Intuition

The goal is to make the sum of the elements in the array greater than or equal to k using the minimum number of operations. The operations you can perform are:

Increase the value of any element by 1.
Duplicate an element and add it to the end of the array.

To approach this problem, think about the operations strategically:

Use the "increase" operation to build up the sum while minimizing operations.
Use the "duplicate" operation to quickly multiply the value in the array once it has reached a certain point.

To find the optimal sequence of operations, you can experiment with how much you increase an element first (a times) before you start duplicating. After increasing a times, compute how many duplications are needed (b times) to reach or exceed the target sum, considering that duplications multiply your sum more rapidly.

Mathematically, if you choose to increase a number a times, the smallest number of duplications needed b is computed using the formula (k + x - 1) // x - 1 where x = a + 1. This formula derives from needing enough copies of an incrementally larger value to reach k. The combination of increments and duplications (a + b) is tried for all possible a, and the minimum result is kept as the answer.

Learn more about Greedy and Math patterns.

Solution Approach

The solution employs a strategy of enumeration, where different combinations of the two operations are tested to find the minimum number of operations needed to make the sum of the array at least k.

Step-by-step Explanation:

Initialize ans to k. This acts as the worst-case starting point where you would have to perform k operations by increasing the element 1 in the array to sum up to k.
Iterate over the potential number of increments a you might use. This is done over the range [0, k], because starting from any value higher than k for a will exceed the necessary operations.
Compute x which represents the value of the element in the array after a increments. This value is given by x = a + 1.
Calculate the smallest number of duplication operations b necessary to reach or exceed k. This uses the formula: [ b = \left\lceil \frac{k}{x} \right\rceil - 1 = \frac{k + x - 1}{x} - 1 ]

This formula calculates the minimum number of duplicates needed of an element valued at x to reach the sum of k.
Update ans with the minimum of the current number of operations (a + b) and the previous ans.
After trying all combinations, ans will hold the minimum number of operations required to achieve the goal.

This solution takes advantage of both increasing and duplicating strategies, cross-verifying each approach's cost-effectiveness by calculating their results in terms of operation count and choosing the minimum.

Overall, this approach efficiently explores possible ways to meet the requirement of sum k through balancing both increasing and duplicating elements.

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Example Walkthrough

Let's consider an example with k = 5. Our goal is to transform the array nums = [1] to have a sum of at least 5 using the minimum number of operations possible.

Initialization: We start by assuming the worst-case scenario where we might need 5 operations if we simply incremented the value from 1 to 5.
Step 2: Try Increasing Then Duplicating: We consider various strategies where we increase elements first for a times and then duplicate as needed.
Iteration for a = 0:
- The value x after 0 increments is 1.
- Calculate duplicates needed: [ b = \left(\frac{5 + 1 - 1}{1}\right) - 1 = 5 - 1 = 4 ]
- Total operations: a + b = 0 + 4 = 4.
Iteration for a = 1:
- The value x after 1 increment is 2.
- Calculate duplicates needed: [ b = \left(\frac{5 + 2 - 1}{2}\right) - 1 = 3 - 1 = 2 ]
- Total operations: a + b = 1 + 2 = 3.
Iteration for a = 2:
- The value x after 2 increments is 3.
- Calculate duplicates needed: [ b = \left(\frac{5 + 3 - 1}{3}\right) - 1 = 2 - 1 = 1 ]
- Total operations: a + b = 2 + 1 = 3.
Iteration for a = 3:
- The value x after 3 increments is 4.
- Calculate duplicates needed: [ b = \left(\frac{5 + 4 - 1}{4}\right) - 1 = 2 - 1 = 1 ]
- Total operations: a + b = 3 + 1 = 4.
Iteration for a = 4:
- The value x after 4 increments is 5.
- Calculate duplicates needed: [ b = \left(\frac{5 + 5 - 1}{5}\right) - 1 = 1 - 1 = 0 ]
- Total operations: a + b = 4 + 0 = 4.

After exploring all possibilities, the minimum number of operations is 3, achieved with a = 1 and b = 2. The optimal strategy is to increase the first element by 1 to make it 2, then duplicate it twice, resulting in the array [2, 2, 2], which sums to 6.

Solution Implementation

1class Solution:
2    def minOperations(self, k: int) -> int:
3        ans = k  # Initialize the answer with k, the maximum number of operations needed in the worst case.
4      
5        # Iterate through possible values of a
6        for a in range(k):
7            x = a + 1  # Increment a by 1 to avoid division by zero and to define x
8            # Calculate b: the number of operations using x as the divisor
9            b = (k + x - 1) // x - 1
10            # Update the answer with the minimum number of operations found so far
11            ans = min(ans, a + b)
12      
13        return ans  # Return the minimum number of operations required
14

1class Solution {
2    public int minOperations(int k) {
3        int ans = k; // Initialize answer with the maximum possible value, which is k.
4      
5        // Iterate through all possible values of 'a' from 0 to k-1.
6        for (int a = 0; a < k; ++a) {
7            int x = a + 1; // Calculate x as a+1, since x must be greater than 0.
8          
9            // Calculate value of b such that b is (k + x - 1) divided by x minus 1.
10            // This effectively partitions k into x-sized groups, and computes the number of remaining elements.
11            int b = (k + x - 1) / x - 1;
12          
13            // Update the answer to the minimum value between current answer and current sum of a and b.
14            ans = Math.min(ans, a + b);
15        }
16      
17        return ans; // Return the minimum operations needed.
18    }
19}
20

1#include <cmath> // Include necessary library for mathematical functions
2
3class Solution {
4public:
5    int minOperations(int k) {
6        int minimumOperations = k; // Initialize minimum operations as k
7      
8        // Iterate through possible values of 'a' from 0 to k-1
9        for (int a = 0; a < k; ++a) {
10            int x = a + 1; // Calculate x as a + 1 
11            // Calculate b based on the formula derived from the problem
12            int b = (k + x - 1) / x - 1;
13          
14            // Update minimumOperations with the minimum of the current value and a + b
15            minimumOperations = std::min(minimumOperations, a + b);
16        }
17      
18        return minimumOperations; // Return the calculated minimum number of operations
19    }
20};
21

1// Function to compute the minimum number of operations required
2function minOperations(k: number): number {
3    // Initialize the answer with the maximum possible value, which is k
4    let ans = k;
5
6    // Iterate through all possible values of 'a' from 0 to k-1
7    for (let a = 0; a < k; ++a) {
8        // Value of x is derived from 'a'
9        const x = a + 1;
10
11        // Calculate 'b' which is the ceiling of k divided by x, minus 1
12        const b = Math.ceil(k / x) - 1;
13
14        // Update the minimum answer
15        ans = Math.min(ans, a + b);
16    }
17
18    // Return the minimum operations found
19    return ans;
20}
21

Time and Space Complexity

The time complexity of the code is O(k). This is because there is a single loop that iterates from 0 to k-1, which results in k iterations. Inside each iteration, the operations performed (such as arithmetic calculations and a comparison with min) are constant-time operations, maintaining the overall complexity of O(k).

The space complexity of the code is O(1). The space usage does not scale with the input size k since only a fixed amount of extra storage is used for the variables ans, a, x, and b. Therefore, the space requirement remains constant, leading to a space complexity of O(1).

Learn more about how to find time and space complexity quickly.

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