Apply Operations to Make Sum of Array Greater Than or Equal to k — LeetCode #3091 (Medium)
Tags: Math, Greedy, Enumeration
Related patterns: Greedy, Enumeration
Brute Force approach
Time complexity: O(n²). Space complexity: O(1).
In the brute force approach, we will try every possible combination of increase and duplicate operations to find the minimum number of operations needed to achieve a sum greater than or equal to k. This method is straightforward but inefficient.
The time complexity is O(n²) because for each increase operation (up to k), we calculate duplicates needed, which can also take linear time in the worst case. The space complexity is O(1) since we only use a few variables to track counts.
- Step 1: Initialize a variable to track the minimum operations needed.
- Step 2: Iterate through all possible numbers of increase operations from 0 to k.
- Step 3: For each number of increases, calculate the resulting sum and determine how many duplicates are needed to reach or exceed k.
- Step 4: Update the minimum operations if the current combination yields a lower count.
For k = 11:
1. Increases = 0: Current Sum = 1, Duplicates Needed = 11, Total Ops = 11
2. Increases = 1: Current Sum = 2, Duplicates Needed = 5, Total Ops = 6
3. Increases = 2: Current Sum = 3, Duplicates Needed = 4, Total Ops = 6
4. Increases = 3: Current Sum = 4, Duplicates Needed = 2, Total Ops = 5
5. Increases = 4: Current Sum = 5, Duplicates Needed = 2, Total Ops = 6
6. Increases = 5: Current Sum = 6, Duplicates Needed = 1, Total Ops = 6
Final Min Ops = 5.
Optimal Solution approach
Time complexity: O(n). Space complexity: O(1).
In the optimal solution, we focus on maximizing the effect of each increase operation before duplicating elements. By calculating how many increases we need first, we can minimize the number of duplicates required to reach or exceed k.
The time complexity is O(n) because we may need to increment the sum up to k. The space complexity is O(1) as we only use a few variables.
- Step 1: Initialize the current sum as 1 and operations as 0.
- Step 2: While the current sum is less than k, increment the current sum by 1 and increment the operations count.
- Step 3: After reaching or exceeding k, calculate how many duplicates are needed based on the current sum.
- Step 4: Return the total operations count.
For k = 11:
1. Current Sum = 1, Operations = 0
2. Increment: Current Sum = 2, Operations = 1
3. Increment: Current Sum = 3, Operations = 2
4. Increment: Current Sum = 4, Operations = 3
5. Increment: Current Sum = 5, Operations = 4
6. Increment: Current Sum = 6, Operations = 5
7. Increment: Current Sum = 7, Operations = 6
8. Increment: Current Sum = 8, Operations = 7
9. Increment: Current Sum = 9, Operations = 8
10. Increment: Current Sum = 10, Operations = 9
11. Increment: Current Sum = 11, Operations = 10
12. Duplicates Needed = 0, Total Ops = 10.
Key Insights
- Maximizing the sum through increases first minimizes the number of duplicates needed.
- Understanding the relationship between increases and duplicates is key to optimizing the solution.
Common Mistakes
- Failing to account for the initial sum of 1 when calculating operations.
- Overcomplicating the problem by not recognizing that increases should be prioritized.
Interview Tips
- Always clarify the problem constraints and initial conditions before jumping into coding.
- Think about edge cases, such as when k is very small or equal to 1.
- Explain your thought process clearly while coding to demonstrate your understanding.
