Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a binary array possible of length n.
Alice and Bob are playing a game that consists of n levels. Some of the levels in the game are impossible to clear while others can always be cleared. In particular, if possible[i] == 0, then the ith level is impossible to clear for both the players. A player gains 1 point on clearing a level and loses 1 point if the player fails to clear it.
At the start of the game, Alice will play some levels in the given order starting from the 0th level, after which Bob will play for the rest of the levels.
Alice wants to know the minimum number of levels she should play to gain more points than Bob, if both players play optimally to maximize their points.
Return the minimum number of levels Alice should play to gain more points. If this is not possible, return -1.
Note that each player must play at least 1 level.
Example 1:
Input: possible = [1,0,1,0]
Output: 1
Explanation:
Let's look at all the levels that Alice can play up to:
Alice must play a minimum of 1 level to gain more points.
Example 2:
Input: possible = [1,1,1,1,1]
Output: 3
Explanation:
Let's look at all the levels that Alice can play up to:
Alice must play a minimum of 3 levels to gain more points.
Example 3:
Input: possible = [0,0]
Output: -1
Explanation:
The only possible way is for both players to play 1 level each. Alice plays level 0 and loses 1 point. Bob plays level 1 and loses 1 point. As both players have equal points, Alice can't gain more points than Bob.
Constraints:
2 <= n == possible.length <= 105possible[i] is either 0 or 1.Problem summary: You are given a binary array possible of length n. Alice and Bob are playing a game that consists of n levels. Some of the levels in the game are impossible to clear while others can always be cleared. In particular, if possible[i] == 0, then the ith level is impossible to clear for both the players. A player gains 1 point on clearing a level and loses 1 point if the player fails to clear it. At the start of the game, Alice will play some levels in the given order starting from the 0th level, after which Bob will play for the rest of the levels. Alice wants to know the minimum number of levels she should play to gain more points than Bob, if both players play optimally to maximize their points. Return the minimum number of levels Alice should play to gain more points. If this is not possible, return -1. Note that each player must play at least 1 level.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[1,0,1,0]
[1,1,1,1,1]
[0,0]
minimum-rounds-to-complete-all-tasks)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3096: Minimum Levels to Gain More Points
class Solution {
public int minimumLevels(int[] possible) {
int s = 0;
for (int x : possible) {
s += x == 0 ? -1 : 1;
}
int t = 0;
for (int i = 1; i < possible.length; ++i) {
t += possible[i - 1] == 0 ? -1 : 1;
if (t > s - t) {
return i;
}
}
return -1;
}
}
// Accepted solution for LeetCode #3096: Minimum Levels to Gain More Points
func minimumLevels(possible []int) int {
s := 0
for _, x := range possible {
if x == 0 {
x = -1
}
s += x
}
t := 0
for i, x := range possible[:len(possible)-1] {
if x == 0 {
x = -1
}
t += x
if t > s-t {
return i + 1
}
}
return -1
}
# Accepted solution for LeetCode #3096: Minimum Levels to Gain More Points
class Solution:
def minimumLevels(self, possible: List[int]) -> int:
s = sum(-1 if x == 0 else 1 for x in possible)
t = 0
for i, x in enumerate(possible[:-1], 1):
t += -1 if x == 0 else 1
if t > s - t:
return i
return -1
// Accepted solution for LeetCode #3096: Minimum Levels to Gain More Points
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #3096: Minimum Levels to Gain More Points
// class Solution {
// public int minimumLevels(int[] possible) {
// int s = 0;
// for (int x : possible) {
// s += x == 0 ? -1 : 1;
// }
// int t = 0;
// for (int i = 1; i < possible.length; ++i) {
// t += possible[i - 1] == 0 ? -1 : 1;
// if (t > s - t) {
// return i;
// }
// }
// return -1;
// }
// }
// Accepted solution for LeetCode #3096: Minimum Levels to Gain More Points
function minimumLevels(possible: number[]): number {
const s = possible.reduce((acc, x) => acc + (x === 0 ? -1 : 1), 0);
let t = 0;
for (let i = 1; i < possible.length; ++i) {
t += possible[i - 1] === 0 ? -1 : 1;
if (t > s - t) {
return i;
}
}
return -1;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.