LeetCode #3477 — EASY

Fruits Into Baskets II

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

The Problem

Problem Statement

You are given two arrays of integers, fruits and baskets, each of length n, where fruits[i] represents the quantity of the i^th type of fruit, and baskets[j] represents the capacity of the j^th basket.

From left to right, place the fruits according to these rules:

Each fruit type must be placed in the leftmost available basket with a capacity greater than or equal to the quantity of that fruit type.
Each basket can hold only one type of fruit.
If a fruit type cannot be placed in any basket, it remains unplaced.

Return the number of fruit types that remain unplaced after all possible allocations are made.

Example 1:

Input: fruits = [4,2,5], baskets = [3,5,4]

Output: 1

Explanation:

fruits[0] = 4 is placed in baskets[1] = 5.
fruits[1] = 2 is placed in baskets[0] = 3.
fruits[2] = 5 cannot be placed in baskets[2] = 4.

Since one fruit type remains unplaced, we return 1.

Example 2:

Input: fruits = [3,6,1], baskets = [6,4,7]

Output: 0

Explanation:

fruits[0] = 3 is placed in baskets[0] = 6.
fruits[1] = 6 cannot be placed in baskets[1] = 4 (insufficient capacity) but can be placed in the next available basket, baskets[2] = 7.
fruits[2] = 1 is placed in baskets[1] = 4.

Since all fruits are successfully placed, we return 0.

Constraints:

n == fruits.length == baskets.length
1 <= n <= 100
1 <= fruits[i], baskets[i] <= 1000

Step 01

Brute Force Baseline

Problem summary: You are given two arrays of integers, fruits and baskets, each of length n, where fruits[i] represents the quantity of the ith type of fruit, and baskets[j] represents the capacity of the jth basket. From left to right, place the fruits according to these rules: Each fruit type must be placed in the leftmost available basket with a capacity greater than or equal to the quantity of that fruit type. Each basket can hold only one type of fruit. If a fruit type cannot be placed in any basket, it remains unplaced. Return the number of fruit types that remain unplaced after all possible allocations are made.

Baseline thinking

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

Pattern signal: Array · Binary Search · Segment Tree

Example 1

[4,2,5]
[3,5,4]

Example 2

[3,6,1]
[6,4,7]

Core Insight

What unlocks the optimal approach

Simulate the operations for each fruit as described

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 #3477: Fruits Into Baskets II
class Solution {
    public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
        int n = fruits.length;
        boolean[] vis = new boolean[n];
        int ans = n;
        for (int x : fruits) {
            for (int i = 0; i < n; ++i) {
                if (baskets[i] >= x && !vis[i]) {
                    vis[i] = true;
                    --ans;
                    break;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3477: Fruits Into Baskets II
func numOfUnplacedFruits(fruits []int, baskets []int) int {
	n := len(fruits)
	ans := n
	vis := make([]bool, n)
	for _, x := range fruits {
		for i, y := range baskets {
			if y >= x && !vis[i] {
				vis[i] = true
				ans--
				break
			}
		}
	}
	return ans
}

# Accepted solution for LeetCode #3477: Fruits Into Baskets II
class Solution:
    def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
        n = len(fruits)
        vis = [False] * n
        ans = n
        for x in fruits:
            for i, y in enumerate(baskets):
                if y >= x and not vis[i]:
                    vis[i] = True
                    ans -= 1
                    break
        return ans

// Accepted solution for LeetCode #3477: Fruits Into Baskets II
// 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 #3477: Fruits Into Baskets II
// class Solution {
//     public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
//         int n = fruits.length;
//         boolean[] vis = new boolean[n];
//         int ans = n;
//         for (int x : fruits) {
//             for (int i = 0; i < n; ++i) {
//                 if (baskets[i] >= x && !vis[i]) {
//                     vis[i] = true;
//                     --ans;
//                     break;
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3477: Fruits Into Baskets II
function numOfUnplacedFruits(fruits: number[], baskets: number[]): number {
    const n = fruits.length;
    const vis: boolean[] = Array(n).fill(false);
    let ans = n;
    for (const x of fruits) {
        for (let i = 0; i < n; ++i) {
            if (baskets[i] >= x && !vis[i]) {
                vis[i] = true;
                --ans;
                break;
            }
        }
    }
    return ans;
}

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)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

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.

BINARY SEARCH

O(log n) time

O(1) space

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).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

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.