LeetCode #3479 — MEDIUM

Fruits Into Baskets III

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
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 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.

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 <= 105
  • 1 <= fruits[i], baskets[i] <= 109
Patterns Used
🔍Modified Binary Search→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
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]

Related Problems

  • Block Placement Queries (block-placement-queries)
Step 02

Core Insight

What unlocks the optimal approach

  • Sort the baskets by the pair of <code>(basket[i], i)</code> in the array.
  • For each fruit from left to right, use binary search to find the first index in the sorted array such that <code>basket[i] >= fruit</code>.
  • Use a segment tree to maintain the smallest original indices where <code>basket[i] >= fruit</code>.
  • When a valid index is found, set the corresponding point to infinity to mark it as used.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. 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 #3479: Fruits Into Baskets III
class SegmentTree {
    int[] nums;
    int[] tr;

    public SegmentTree(int[] nums) {
        this.nums = nums;
        int n = nums.length;
        this.tr = new int[n << 2];
        build(1, 1, n);
    }

    public void build(int u, int l, int r) {
        if (l == r) {
            tr[u] = nums[l - 1];
            return;
        }
        int mid = (l + r) >> 1;
        build(u << 1, l, mid);
        build(u << 1 | 1, mid + 1, r);
        pushup(u);
    }

    public void modify(int u, int l, int r, int i, int v) {
        if (l == r) {
            tr[u] = v;
            return;
        }
        int mid = (l + r) >> 1;
        if (i <= mid) {
            modify(u << 1, l, mid, i, v);
        } else {
            modify(u << 1 | 1, mid + 1, r, i, v);
        }
        pushup(u);
    }

    public int query(int u, int l, int r, int v) {
        if (tr[u] < v) {
            return -1;
        }
        if (l == r) {
            return l;
        }
        int mid = (l + r) >> 1;
        if (tr[u << 1] >= v) {
            return query(u << 1, l, mid, v);
        }
        return query(u << 1 | 1, mid + 1, r, v);
    }

    public void pushup(int u) {
        tr[u] = Math.max(tr[u << 1], tr[u << 1 | 1]);
    }
}

class Solution {
    public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
        SegmentTree tree = new SegmentTree(baskets);
        int n = baskets.length;
        int ans = 0;
        for (int x : fruits) {
            int i = tree.query(1, 1, n, x);
            if (i < 0) {
                ans++;
            } else {
                tree.modify(1, 1, n, i, 0);
            }
        }
        return ans;
    }
}
Step 06

Interactive Study Demo

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

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log 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.

Related Problems
#315Count of Smaller Numbers After Selfhard→#327Count of Range Sumhard→#363Max Sum of Rectangle No Larger Than Khard→#456132 Patternmedium→#493Reverse Pairshard→