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.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given a 0-indexed array usageLimits of length n.
Your task is to create groups using numbers from 0 to n - 1, ensuring that each number, i, is used no more than usageLimits[i] times in total across all groups. You must also satisfy the following conditions:
Return an integer denoting the maximum number of groups you can create while satisfying these conditions.
Example 1:
Input: usageLimits = [1,2,5]
Output: 3
Explanation: In this example, we can use 0 at most once, 1 at most twice, and 2 at most five times.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number [2].
Group 2 contains the numbers [1,2].
Group 3 contains the numbers [0,1,2].
It can be shown that the maximum number of groups is 3.
So, the output is 3.
Example 2:
Input: usageLimits = [2,1,2]
Output: 2
Explanation: In this example, we can use 0 at most twice, 1 at most once, and 2 at most twice.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number [0].
Group 2 contains the numbers [1,2].
It can be shown that the maximum number of groups is 2.
So, the output is 2.
Example 3:
Input: usageLimits = [1,1]
Output: 1
Explanation: In this example, we can use both 0 and 1 at most once.
One way of creating the maximum number of groups while satisfying the conditions is:
Group 1 contains the number [0].
It can be shown that the maximum number of groups is 1.
So, the output is 1.
Constraints:
1 <= usageLimits.length <= 1051 <= usageLimits[i] <= 109Problem summary: You are given a 0-indexed array usageLimits of length n. Your task is to create groups using numbers from 0 to n - 1, ensuring that each number, i, is used no more than usageLimits[i] times in total across all groups. You must also satisfy the following conditions: Each group must consist of distinct numbers, meaning that no duplicate numbers are allowed within a single group. Each group (except the first one) must have a length strictly greater than the previous group. Return an integer denoting the maximum number of groups you can create while satisfying these conditions.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Math · Binary Search · Greedy
[1,2,5]
[2,1,2]
[1,1]
group-the-people-given-the-group-size-they-belong-to)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2790: Maximum Number of Groups With Increasing Length
class Solution {
public int maxIncreasingGroups(List<Integer> usageLimits) {
Collections.sort(usageLimits);
int k = 0;
long s = 0;
for (int x : usageLimits) {
s += x;
if (s > k) {
++k;
s -= k;
}
}
return k;
}
}
// Accepted solution for LeetCode #2790: Maximum Number of Groups With Increasing Length
func maxIncreasingGroups(usageLimits []int) int {
sort.Ints(usageLimits)
s, k := 0, 0
for _, x := range usageLimits {
s += x
if s > k {
k++
s -= k
}
}
return k
}
# Accepted solution for LeetCode #2790: Maximum Number of Groups With Increasing Length
class Solution:
def maxIncreasingGroups(self, usageLimits: List[int]) -> int:
usageLimits.sort()
k, n = 0, len(usageLimits)
for i in range(n):
if usageLimits[i] > k:
k += 1
usageLimits[i] -= k
if i + 1 < n:
usageLimits[i + 1] += usageLimits[i]
return k
// Accepted solution for LeetCode #2790: Maximum Number of Groups With Increasing Length
// 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 #2790: Maximum Number of Groups With Increasing Length
// class Solution {
// public int maxIncreasingGroups(List<Integer> usageLimits) {
// Collections.sort(usageLimits);
// int k = 0;
// long s = 0;
// for (int x : usageLimits) {
// s += x;
// if (s > k) {
// ++k;
// s -= k;
// }
// }
// return k;
// }
// }
// Accepted solution for LeetCode #2790: Maximum Number of Groups With Increasing Length
function maxIncreasingGroups(usageLimits: number[]): number {
usageLimits.sort((a, b) => a - b);
let k = 0;
let s = 0;
for (const x of usageLimits) {
s += x;
if (s > k) {
++k;
s -= k;
}
}
return k;
}
Use this to step through a reusable interview workflow for this problem.
Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.
Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).
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.
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: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.
Usually fails on: Two-element ranges never converge.
Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.
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.