LeetCode #2591 — EASY

Distribute Money to Maximum Children

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

Problem Statement

You are given an integer money denoting the amount of money (in dollars) that you have and another integer children denoting the number of children that you must distribute the money to.

You have to distribute the money according to the following rules:

All money must be distributed.
Everyone must receive at least 1 dollar.
Nobody receives 4 dollars.

Return the maximum number of children who may receive exactly 8 dollars if you distribute the money according to the aforementioned rules. If there is no way to distribute the money, return -1.

Example 1:

Input: money = 20, children = 3
Output: 1
Explanation: 
The maximum number of children with 8 dollars will be 1. One of the ways to distribute the money is:
- 8 dollars to the first child.
- 9 dollars to the second child. 
- 3 dollars to the third child.
It can be proven that no distribution exists such that number of children getting 8 dollars is greater than 1.

Example 2:

Input: money = 16, children = 2
Output: 2
Explanation: Each child can be given 8 dollars.

Constraints:

1 <= money <= 200
2 <= children <= 30

Step 01

Brute Force Baseline

Problem summary: You are given an integer money denoting the amount of money (in dollars) that you have and another integer children denoting the number of children that you must distribute the money to. You have to distribute the money according to the following rules: All money must be distributed. Everyone must receive at least 1 dollar. Nobody receives 4 dollars. Return the maximum number of children who may receive exactly 8 dollars if you distribute the money according to the aforementioned rules. If there is no way to distribute the money, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Greedy

Example 1

20
3

Example 2

16
2

Core Insight

What unlocks the optimal approach

Can we distribute the money according to the rules if we give 'k' children exactly 8 dollars?
Brute force to find the largest possible value of k, or return -1 if there doesn’t exist any such k.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2591: Distribute Money to Maximum Children
class Solution {
    public int distMoney(int money, int children) {
        if (money < children) {
            return -1;
        }
        if (money > 8 * children) {
            return children - 1;
        }
        if (money == 8 * children - 4) {
            return children - 2;
        }
        // money-8x >= children-x, x <= (money-children)/7
        return (money - children) / 7;
    }
}

// Accepted solution for LeetCode #2591: Distribute Money to Maximum Children
func distMoney(money int, children int) int {
	if money < children {
		return -1
	}
	if money > 8*children {
		return children - 1
	}
	if money == 8*children-4 {
		return children - 2
	}
	// money-8x >= children-x, x <= (money-children)/7
	return (money - children) / 7
}

# Accepted solution for LeetCode #2591: Distribute Money to Maximum Children
class Solution:
    def distMoney(self, money: int, children: int) -> int:
        if money < children:
            return -1
        if money > 8 * children:
            return children - 1
        if money == 8 * children - 4:
            return children - 2
        # money-8x >= children-x, x <= (money-children)/7
        return (money - children) // 7

// Accepted solution for LeetCode #2591: Distribute Money to Maximum Children
impl Solution {
    pub fn dist_money(money: i32, children: i32) -> i32 {
        if money < children {
            return -1;
        }

        if money > children * 8 {
            return children - 1;
        }

        if money == children * 8 - 4 {
            return children - 2;
        }

        (money - children) / 7
    }
}

// Accepted solution for LeetCode #2591: Distribute Money to Maximum Children
function distMoney(money: number, children: number): number {
    if (money < children) {
        return -1;
    }
    if (money > 8 * children) {
        return children - 1;
    }
    if (money === 8 * children - 4) {
        return children - 2;
    }
    return Math.floor((money - children) / 7);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

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

O(n log n) time

O(1) 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.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Using greedy without proof

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.