LeetCode #638 — MEDIUM

Shopping Offers

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

The Problem

Problem Statement

In LeetCode Store, there are n items to sell. Each item has a price. However, there are some special offers, and a special offer consists of one or more different kinds of items with a sale price.

You are given an integer array price where price[i] is the price of the i^th item, and an integer array needs where needs[i] is the number of pieces of the i^th item you want to buy.

You are also given an array special where special[i] is of size n + 1 where special[i][j] is the number of pieces of the j^th item in the i^th offer and special[i][n] (i.e., the last integer in the array) is the price of the i^th offer.

Return the lowest price you have to pay for exactly certain items as given, where you could make optimal use of the special offers. You are not allowed to buy more items than you want, even if that would lower the overall price. You could use any of the special offers as many times as you want.

Example 1:

Input: price = [2,5], special = [[3,0,5],[1,2,10]], needs = [3,2]
Output: 14
Explanation: There are two kinds of items, A and B. Their prices are $2 and $5 respectively. 
In special offer 1, you can pay $5 for 3A and 0B
In special offer 2, you can pay $10 for 1A and 2B. 
You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.

Example 2:

Input: price = [2,3,4], special = [[1,1,0,4],[2,2,1,9]], needs = [1,2,1]
Output: 11
Explanation: The price of A is $2, and $3 for B, $4 for C. 
You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C. 
You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C. 
You cannot add more items, though only $9 for 2A ,2B and 1C.

Constraints:

n == price.length == needs.length
1 <= n <= 6
0 <= price[i], needs[i] <= 10
1 <= special.length <= 100
special[i].length == n + 1
0 <= special[i][j] <= 50
The input is generated that at least one of special[i][j] is non-zero for 0 <= j <= n - 1.

Step 01

Brute Force Baseline

Problem summary: In LeetCode Store, there are n items to sell. Each item has a price. However, there are some special offers, and a special offer consists of one or more different kinds of items with a sale price. You are given an integer array price where price[i] is the price of the ith item, and an integer array needs where needs[i] is the number of pieces of the ith item you want to buy. You are also given an array special where special[i] is of size n + 1 where special[i][j] is the number of pieces of the jth item in the ith offer and special[i][n] (i.e., the last integer in the array) is the price of the ith offer. Return the lowest price you have to pay for exactly certain items as given, where you could make optimal use of the special offers. You are not allowed to buy more items than you want, even if that would lower the overall price. You could use any of the special offers as many times as

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Backtracking · Bit Manipulation

Example 1

[2,5]
[[3,0,5],[1,2,10]]
[3,2]

Example 2

[2,3,4]
[[1,1,0,4],[2,2,1,9]]
[1,2,1]

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 #638: Shopping Offers
class Solution {
    private final int bits = 4;
    private int n;
    private List<Integer> price;
    private List<List<Integer>> special;
    private Map<Integer, Integer> f = new HashMap<>();

    public int shoppingOffers(
        List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
        n = needs.size();
        this.price = price;
        this.special = special;
        int mask = 0;
        for (int i = 0; i < n; ++i) {
            mask |= needs.get(i) << (i * bits);
        }
        return dfs(mask);
    }

    private int dfs(int cur) {
        if (f.containsKey(cur)) {
            return f.get(cur);
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += price.get(i) * (cur >> (i * bits) & 0xf);
        }
        for (List<Integer> offer : special) {
            int nxt = cur;
            boolean ok = true;
            for (int j = 0; j < n; ++j) {
                if ((cur >> (j * bits) & 0xf) < offer.get(j)) {
                    ok = false;
                    break;
                }
                nxt -= offer.get(j) << (j * bits);
            }
            if (ok) {
                ans = Math.min(ans, offer.get(n) + dfs(nxt));
            }
        }
        f.put(cur, ans);
        return ans;
    }
}

// Accepted solution for LeetCode #638: Shopping Offers
func shoppingOffers(price []int, special [][]int, needs []int) int {
	const bits = 4
	n := len(needs)
	f := make(map[int]int)
	mask := 0
	for i, need := range needs {
		mask |= need << (i * bits)
	}

	var dfs func(int) int
	dfs = func(cur int) int {
		if v, ok := f[cur]; ok {
			return v
		}
		ans := 0
		for i := 0; i < n; i++ {
			ans += price[i] * ((cur >> (i * bits)) & 0xf)
		}
		for _, offer := range special {
			nxt := cur
			ok := true
			for j := 0; j < n; j++ {
				if ((cur >> (j * bits)) & 0xf) < offer[j] {
					ok = false
					break
				}
				nxt -= offer[j] << (j * bits)
			}
			if ok {
				ans = min(ans, offer[n]+dfs(nxt))
			}
		}
		f[cur] = ans
		return ans
	}

	return dfs(mask)
}

# Accepted solution for LeetCode #638: Shopping Offers
class Solution:
    def shoppingOffers(
        self, price: List[int], special: List[List[int]], needs: List[int]
    ) -> int:
        @cache
        def dfs(cur: int) -> int:
            ans = sum(p * (cur >> (i * bits) & 0xF) for i, p in enumerate(price))
            for offer in special:
                nxt = cur
                for j in range(len(needs)):
                    if (cur >> (j * bits) & 0xF) < offer[j]:
                        break
                    nxt -= offer[j] << (j * bits)
                else:
                    ans = min(ans, offer[-1] + dfs(nxt))
            return ans

        bits, mask = 4, 0
        for i, need in enumerate(needs):
            mask |= need << i * bits
        return dfs(mask)

// Accepted solution for LeetCode #638: Shopping Offers
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn shopping_offers(price: Vec<i32>, special: Vec<Vec<i32>>, needs: Vec<i32>) -> i32 {
        let n = price.len();
        let m = special.len();
        let mut memo: HashMap<Vec<i32>, i32> = HashMap::new();
        memo.insert(vec![0; n], 0);
        Self::dfs(needs, &mut memo, &price, &special, n, m)
    }
    fn dfs(
        needs: Vec<i32>,
        memo: &mut HashMap<Vec<i32>, i32>,
        price: &[i32],
        special: &[Vec<i32>],
        n: usize,
        m: usize,
    ) -> i32 {
        if let Some(&res) = memo.get(&needs) {
            return res;
        }
        let mut res: i32 = needs.iter().zip(price.iter()).map(|(a, b)| a * b).sum();
        'special: for j in 0..m {
            for i in 0..n {
                if special[j][i] > needs[i] {
                    continue 'special;
                }
            }
            let mut needs = needs.to_vec();
            for i in 0..n {
                needs[i] -= special[j][i];
            }
            res = res.min(Self::dfs(needs, memo, price, special, n, m) + special[j][n]);
        }
        memo.insert(needs, res);
        res
    }
}

#[test]
fn test() {
    let price = vec![2, 5];
    let special = vec_vec_i32![[3, 0, 5], [1, 2, 10]];
    let needs = vec![3, 2];
    let res = 14;
    assert_eq!(Solution::shopping_offers(price, special, needs), res);
    let price = vec![2, 3, 4];
    let special = vec_vec_i32![[1, 1, 0, 4], [2, 2, 1, 9]];
    let needs = vec![1, 2, 1];
    let res = 11;
    assert_eq!(Solution::shopping_offers(price, special, needs), res);
}

// Accepted solution for LeetCode #638: Shopping Offers
function shoppingOffers(price: number[], special: number[][], needs: number[]): number {
    const bits = 4;
    const n = needs.length;
    const f: Map<number, number> = new Map();

    let mask = 0;
    for (let i = 0; i < n; i++) {
        mask |= needs[i] << (i * bits);
    }

    const dfs = (cur: number): number => {
        if (f.has(cur)) {
            return f.get(cur)!;
        }
        let ans = 0;
        for (let i = 0; i < n; i++) {
            ans += price[i] * ((cur >> (i * bits)) & 0xf);
        }
        for (const offer of special) {
            let nxt = cur;
            let ok = true;
            for (let j = 0; j < n; j++) {
                if (((cur >> (j * bits)) & 0xf) < offer[j]) {
                    ok = false;
                    break;
                }
                nxt -= offer[j] << (j * bits);
            }
            if (ok) {
                ans = Math.min(ans, offer[n] + dfs(nxt));
            }
        }
        f.set(cur, ans);
        return ans;
    };

    return dfs(mask);
}

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 × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.