LeetCode #455 — EASY

Assign Cookies

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

The Problem

Problem Statement

Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.

Each child i has a greed factor g[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s[j]. If s[j] >= g[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

Example 1:

Input: g = [1,2,3], s = [1,1]
Output: 1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3. 
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.

Example 2:

Input: g = [1,2], s = [1,2,3]
Output: 2
Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2. 
You have 3 cookies and their sizes are big enough to gratify all of the children, 
You need to output 2.

Constraints:

1 <= g.length <= 3 * 10⁴
0 <= s.length <= 3 * 10⁴
1 <= g[i], s[j] <= 2³¹ - 1

Note: This question is the same as 2410: Maximum Matching of Players With Trainers.

Step 01

Brute Force Baseline

Problem summary: Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie. Each child i has a greed factor g[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s[j]. If s[j] >= g[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.

Baseline thinking

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

Pattern signal: Array · Two Pointers · Greedy

Example 1

[1,2,3]
[1,1]

Example 2

[1,2]
[1,2,3]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #455: Assign Cookies
class Solution {
    public int findContentChildren(int[] g, int[] s) {
        Arrays.sort(g);
        Arrays.sort(s);
        int m = g.length;
        int n = s.length;
        for (int i = 0, j = 0; i < m; ++i) {
            while (j < n && s[j] < g[i]) {
                ++j;
            }
            if (j++ >= n) {
                return i;
            }
        }
        return m;
    }
}

// Accepted solution for LeetCode #455: Assign Cookies
func findContentChildren(g []int, s []int) int {
	sort.Ints(g)
	sort.Ints(s)
	j := 0
	for i, x := range g {
		for j < len(s) && s[j] < x {
			j++
		}
		if j >= len(s) {
			return i
		}
		j++
	}
	return len(g)
}

# Accepted solution for LeetCode #455: Assign Cookies
class Solution:
    def findContentChildren(self, g: List[int], s: List[int]) -> int:
        g.sort()
        s.sort()
        j = 0
        for i, x in enumerate(g):
            while j < len(s) and s[j] < g[i]:
                j += 1
            if j >= len(s):
                return i
            j += 1
        return len(g)

// Accepted solution for LeetCode #455: Assign Cookies
struct Solution;

impl Solution {
    fn find_content_children(mut g: Vec<i32>, mut s: Vec<i32>) -> i32 {
        g.sort_unstable();
        s.sort_unstable();
        let mut i = 0;
        let mut j = 0;
        while i < g.len() && j < s.len() {
            if g[i] <= s[j] {
                i += 1;
                j += 1;
            } else {
                while j < s.len() && s[j] < g[i] {
                    j += 1;
                }
            }
        }
        i as i32
    }
}

#[test]
fn test() {
    let g = vec![1, 2, 3];
    let s = vec![1, 1];
    let res = 1;
    assert_eq!(Solution::find_content_children(g, s), res);
    let g = vec![1, 2];
    let s = vec![1, 2, 3];
    let res = 2;
    assert_eq!(Solution::find_content_children(g, s), res);
}

// Accepted solution for LeetCode #455: Assign Cookies
function findContentChildren(g: number[], s: number[]): number {
    g.sort((a, b) => a - b);
    s.sort((a, b) => a - b);
    const m = g.length;
    const n = s.length;
    for (let i = 0, j = 0; i < m; ++i) {
        while (j < n && s[j] < g[i]) {
            ++j;
        }
        if (j++ >= n) {
            return i;
        }
    }
    return m;
}

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(m × log m + n × log n)

Space

O(log m + log n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

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.