Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Move from brute-force thinking to an efficient approach using math strategy.
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):
1.Return the minimum number of operations required to make the sum of elements of the final array greater than or equal to k.
Example 1:
Input: k = 11
Output: 5
Explanation:
We can do the following operations on the array nums = [1]:
1 three times. The resulting array is nums = [4].nums = [4,4,4].The sum of the final array is 4 + 4 + 4 = 12 which is greater than or equal to k = 11.
The total number of operations performed is 3 + 2 = 5.
Example 2:
Input: k = 1
Output: 0
Explanation:
The sum of the original array is already greater than or equal to 1, so no operations are needed.
Constraints:
1 <= k <= 105Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Greedy
11
1
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3091: Apply Operations to Make Sum of Array Greater Than or Equal to k
class Solution {
public int minOperations(int k) {
int ans = k;
for (int a = 0; a < k; ++a) {
int x = a + 1;
int b = (k + x - 1) / x - 1;
ans = Math.min(ans, a + b);
}
return ans;
}
}
// Accepted solution for LeetCode #3091: Apply Operations to Make Sum of Array Greater Than or Equal to k
func minOperations(k int) int {
ans := k
for a := 0; a < k; a++ {
x := a + 1
b := (k+x-1)/x - 1
ans = min(ans, a+b)
}
return ans
}
# Accepted solution for LeetCode #3091: Apply Operations to Make Sum of Array Greater Than or Equal to k
class Solution:
def minOperations(self, k: int) -> int:
ans = k
for a in range(k):
x = a + 1
b = (k + x - 1) // x - 1
ans = min(ans, a + b)
return ans
// Accepted solution for LeetCode #3091: Apply Operations to Make Sum of Array Greater Than or Equal to k
impl Solution {
pub fn min_operations(k: i32) -> i32 {
let mut ans = k;
for a in 0..k {
let x = a + 1;
let b = (k + x - 1) / x - 1;
ans = ans.min(a + b);
}
ans
}
}
// Accepted solution for LeetCode #3091: Apply Operations to Make Sum of Array Greater Than or Equal to k
function minOperations(k: number): number {
let ans = k;
for (let a = 0; a < k; ++a) {
const x = a + 1;
const b = Math.ceil(k / x) - 1;
ans = Math.min(ans, a + b);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.
Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.