LeetCode #2944 — MEDIUM

Minimum Number of Coins for Fruits

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

The Problem

Problem Statement

You are given an 0-indexed integer array prices where prices[i] denotes the number of coins needed to purchase the (i + 1)^th fruit.

The fruit market has the following reward for each fruit:

If you purchase the (i + 1)^th fruit at prices[i] coins, you can get any number of the next i fruits for free.

Note that even if you can take fruit j for free, you can still purchase it for prices[j - 1] coins to receive its reward.

Return the minimum number of coins needed to acquire all the fruits.

Example 1:

Input: prices = [3,1,2]

Output: 4

Explanation:

Purchase the 1^st fruit with prices[0] = 3 coins, you are allowed to take the 2^nd fruit for free.
Purchase the 2^nd fruit with prices[1] = 1 coin, you are allowed to take the 3^rd fruit for free.
Take the 3^rd fruit for free.

Note that even though you could take the 2^nd fruit for free as a reward of buying 1^st fruit, you purchase it to receive its reward, which is more optimal.

Example 2:

Input: prices = [1,10,1,1]

Output: 2

Explanation:

Purchase the 1^st fruit with prices[0] = 1 coin, you are allowed to take the 2^nd fruit for free.
Take the 2^nd fruit for free.
Purchase the 3^rd fruit for prices[2] = 1 coin, you are allowed to take the 4^th fruit for free.
Take the 4^t^h fruit for free.

Example 3:

Input: prices = [26,18,6,12,49,7,45,45]

Output: 39

Explanation:

Purchase the 1^st fruit with prices[0] = 26 coin, you are allowed to take the 2^nd fruit for free.
Take the 2^nd fruit for free.
Purchase the 3^rd fruit for prices[2] = 6 coin, you are allowed to take the 4^th, 5^th and 6^th (the next three) fruits for free.
Take the 4^t^h fruit for free.
Take the 5^t^h fruit for free.
Purchase the 6^th fruit with prices[5] = 7 coin, you are allowed to take the 8^th and 9^th fruit for free.
Take the 7^t^h fruit for free.
Take the 8^t^h fruit for free.

Note that even though you could take the 6^th fruit for free as a reward of buying 3^rd fruit, you purchase it to receive its reward, which is more optimal.

Constraints:

1 <= prices.length <= 1000
1 <= prices[i] <= 10⁵

Step 01

Brute Force Baseline

Problem summary: You are given an 0-indexed integer array prices where prices[i] denotes the number of coins needed to purchase the (i + 1)th fruit. The fruit market has the following reward for each fruit: If you purchase the (i + 1)th fruit at prices[i] coins, you can get any number of the next i fruits for free. Note that even if you can take fruit j for free, you can still purchase it for prices[j - 1] coins to receive its reward. Return the minimum number of coins needed to acquire all the fruits.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming · Monotonic Queue

Example 1

[3,1,2]

Example 2

[1,10,1,1]

Example 3

[26,18,6,12,49,7,45,45]

Step 02

Core Insight

What unlocks the optimal approach

The intended solution uses Dynamic Programming.
Let <code>dp[i]</code> denote the minimum number of coins, such that we bought <code>i<sup>th</sup></code> fruit and acquired all the fruits in the range <code>[i...n]</code>.
<code>dp[i] = min(dp[i], dp[j] + prices[i]) </code>, where <code>j</code> is in the range <code>[i + 1, i + 1 + i]</code>.

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 #2944: Minimum Number of Coins for Fruits
class Solution {
    private int[] prices;
    private int[] f;
    private int n;

    public int minimumCoins(int[] prices) {
        n = prices.length;
        f = new int[n + 1];
        this.prices = prices;
        return dfs(1);
    }

    private int dfs(int i) {
        if (i * 2 >= n) {
            return prices[i - 1];
        }
        if (f[i] == 0) {
            f[i] = 1 << 30;
            for (int j = i + 1; j <= i * 2 + 1; ++j) {
                f[i] = Math.min(f[i], prices[i - 1] + dfs(j));
            }
        }
        return f[i];
    }
}

// Accepted solution for LeetCode #2944: Minimum Number of Coins for Fruits
func minimumCoins(prices []int) int {
	n := len(prices)
	f := make([]int, n+1)
	var dfs func(int) int
	dfs = func(i int) int {
		if i*2 >= n {
			return prices[i-1]
		}
		if f[i] == 0 {
			f[i] = 1 << 30
			for j := i + 1; j <= i*2+1; j++ {
				f[i] = min(f[i], dfs(j)+prices[i-1])
			}
		}
		return f[i]
	}
	return dfs(1)
}

# Accepted solution for LeetCode #2944: Minimum Number of Coins for Fruits
class Solution:
    def minimumCoins(self, prices: List[int]) -> int:
        @cache
        def dfs(i: int) -> int:
            if i * 2 >= len(prices):
                return prices[i - 1]
            return prices[i - 1] + min(dfs(j) for j in range(i + 1, i * 2 + 2))

        return dfs(1)

// Accepted solution for LeetCode #2944: Minimum Number of Coins for Fruits
/**
 * [2944] Minimum Number of Coins for Fruits
 */
pub struct Solution {}

// submission codes start here
use std::collections::VecDeque;

impl Solution {
    pub fn minimum_coins(prices: Vec<i32>) -> i32 {
        let n = prices.len();
        // (滑动窗口的右侧, dp[i])
        let mut queue = VecDeque::from([(n + 1, 0)]);

        for i in (1..=n).rev() {
            while queue.back().unwrap().0 > i * 2 + 1 {
                queue.pop_back();
            }

            let dp = prices[i - 1] + queue.back().unwrap().1;
            while dp <= queue.front().unwrap().1 {
                queue.pop_front();
            }

            queue.push_front((i, dp));
        }

        queue.front().unwrap().1
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2944() {
        assert_eq!(4, Solution::minimum_coins(vec![3, 1, 2]));
        assert_eq!(2, Solution::minimum_coins(vec![1, 10, 1, 1]));
        assert_eq!(
            39,
            Solution::minimum_coins(vec![26, 18, 6, 12, 49, 7, 45, 45])
        );
    }
}

// Accepted solution for LeetCode #2944: Minimum Number of Coins for Fruits
function minimumCoins(prices: number[]): number {
    const n = prices.length;
    const f: number[] = Array(n + 1).fill(0);
    const dfs = (i: number): number => {
        if (i * 2 >= n) {
            return prices[i - 1];
        }
        if (f[i] === 0) {
            f[i] = 1 << 30;
            for (let j = i + 1; j <= i * 2 + 1; ++j) {
                f[i] = Math.min(f[i], prices[i - 1] + dfs(j));
            }
        }
        return f[i];
    };
    return dfs(1);
}

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^2)

Space

O(n)

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.