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 playing a game with integers. You start with the integer 1 and you want to reach the integer target.
In one move, you can either:
x = x + 1).x = 2 * x).You can use the increment operation any number of times, however, you can only use the double operation at most maxDoubles times.
Given the two integers target and maxDoubles, return the minimum number of moves needed to reach target starting with 1.
Example 1:
Input: target = 5, maxDoubles = 0 Output: 4 Explanation: Keep incrementing by 1 until you reach target.
Example 2:
Input: target = 19, maxDoubles = 2 Output: 7 Explanation: Initially, x = 1 Increment 3 times so x = 4 Double once so x = 8 Increment once so x = 9 Double again so x = 18 Increment once so x = 19
Example 3:
Input: target = 10, maxDoubles = 4 Output: 4 Explanation: Initially, x = 1 Increment once so x = 2 Double once so x = 4 Increment once so x = 5 Double again so x = 10
Constraints:
1 <= target <= 1090 <= maxDoubles <= 100Problem summary: You are playing a game with integers. You start with the integer 1 and you want to reach the integer target. In one move, you can either: Increment the current integer by one (i.e., x = x + 1). Double the current integer (i.e., x = 2 * x). You can use the increment operation any number of times, however, you can only use the double operation at most maxDoubles times. Given the two integers target and maxDoubles, return the minimum number of moves needed to reach target starting with 1.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Greedy
5 0
19 2
10 4
number-of-steps-to-reduce-a-number-to-zero)number-of-steps-to-reduce-a-number-in-binary-representation-to-one)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2139: Minimum Moves to Reach Target Score
class Solution {
public int minMoves(int target, int maxDoubles) {
if (target == 1) {
return 0;
}
if (maxDoubles == 0) {
return target - 1;
}
if (target % 2 == 0 && maxDoubles > 0) {
return 1 + minMoves(target >> 1, maxDoubles - 1);
}
return 1 + minMoves(target - 1, maxDoubles);
}
}
// Accepted solution for LeetCode #2139: Minimum Moves to Reach Target Score
func minMoves(target int, maxDoubles int) int {
if target == 1 {
return 0
}
if maxDoubles == 0 {
return target - 1
}
if target%2 == 0 && maxDoubles > 0 {
return 1 + minMoves(target>>1, maxDoubles-1)
}
return 1 + minMoves(target-1, maxDoubles)
}
# Accepted solution for LeetCode #2139: Minimum Moves to Reach Target Score
class Solution:
def minMoves(self, target: int, maxDoubles: int) -> int:
if target == 1:
return 0
if maxDoubles == 0:
return target - 1
if target % 2 == 0 and maxDoubles:
return 1 + self.minMoves(target >> 1, maxDoubles - 1)
return 1 + self.minMoves(target - 1, maxDoubles)
// Accepted solution for LeetCode #2139: Minimum Moves to Reach Target Score
/**
* [2139] Minimum Moves to Reach Target Score
*
* You are playing a game with integers. You start with the integer 1 and you want to reach the integer target.
* In one move, you can either:
*
* Increment the current integer by one (i.e., x = x + 1).
* Double the current integer (i.e., x = 2 * x).
*
* You can use the increment operation any number of times, however, you can only use the double operation at most maxDoubles times.
* Given the two integers target and maxDoubles, return the minimum number of moves needed to reach target starting with 1.
*
* Example 1:
*
* Input: target = 5, maxDoubles = 0
* Output: 4
* Explanation: Keep incrementing by 1 until you reach target.
*
* Example 2:
*
* Input: target = 19, maxDoubles = 2
* Output: 7
* Explanation: Initially, x = 1
* Increment 3 times so x = 4
* Double once so x = 8
* Increment once so x = 9
* Double again so x = 18
* Increment once so x = 19
*
* Example 3:
*
* Input: target = 10, maxDoubles = 4
* Output: 4
* Explanation: Initially, x = 1
* Increment once so x = 2
* Double once so x = 4
* Increment once so x = 5
* Double again so x = 10
*
*
* Constraints:
*
* 1 <= target <= 10^9
* 0 <= maxDoubles <= 100
*
*/
pub struct Solution {}
// problem: https://leetcode.com/problems/minimum-moves-to-reach-target-score/
// discuss: https://leetcode.com/problems/minimum-moves-to-reach-target-score/discuss/?currentPage=1&orderBy=most_votes&query=
// submission codes start here
impl Solution {
pub fn min_moves(target: i32, max_doubles: i32) -> i32 {
0
}
}
// submission codes end
#[cfg(test)]
mod tests {
use std::result;
use super::*;
#[test]
#[ignore]
fn test_2139_example_1() {
let target = 5;
let max_doubles = 0;
let result = 4;
assert_eq!(Solution::min_moves(target, max_doubles), result);
}
#[test]
#[ignore]
fn test_2139_example_2() {
let target = 19;
let max_doubles = 2;
let result = 7;
assert_eq!(Solution::min_moves(target, max_doubles), result);
}
#[test]
#[ignore]
fn test_2139_example_3() {
let target = 10;
let max_doubles = 4;
let result = 4;
assert_eq!(Solution::min_moves(target, max_doubles), result);
}
}
// Accepted solution for LeetCode #2139: Minimum Moves to Reach Target Score
function minMoves(target: number, maxDoubles: number): number {
if (target === 1) {
return 0;
}
if (maxDoubles === 0) {
return target - 1;
}
if (target % 2 === 0 && maxDoubles) {
return 1 + minMoves(target >> 1, maxDoubles - 1);
}
return 1 + minMoves(target - 1, maxDoubles);
}
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.