LeetCode #1553 — HARD

Minimum Number of Days to Eat N Oranges

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

There are n oranges in the kitchen and you decided to eat some of these oranges every day as follows:

Eat one orange.
If the number of remaining oranges n is divisible by 2 then you can eat n / 2 oranges.
If the number of remaining oranges n is divisible by 3 then you can eat 2 * (n / 3) oranges.

You can only choose one of the actions per day.

Given the integer n, return the minimum number of days to eat n oranges.

Example 1:

Input: n = 10
Output: 4
Explanation: You have 10 oranges.
Day 1: Eat 1 orange,  10 - 1 = 9.  
Day 2: Eat 6 oranges, 9 - 2*(9/3) = 9 - 6 = 3. (Since 9 is divisible by 3)
Day 3: Eat 2 oranges, 3 - 2*(3/3) = 3 - 2 = 1. 
Day 4: Eat the last orange  1 - 1  = 0.
You need at least 4 days to eat the 10 oranges.

Example 2:

Input: n = 6
Output: 3
Explanation: You have 6 oranges.
Day 1: Eat 3 oranges, 6 - 6/2 = 6 - 3 = 3. (Since 6 is divisible by 2).
Day 2: Eat 2 oranges, 3 - 2*(3/3) = 3 - 2 = 1. (Since 3 is divisible by 3)
Day 3: Eat the last orange  1 - 1  = 0.
You need at least 3 days to eat the 6 oranges.

Constraints:

1 <= n <= 2 * 10⁹

Step 01

Brute Force Baseline

Problem summary: There are n oranges in the kitchen and you decided to eat some of these oranges every day as follows: Eat one orange. If the number of remaining oranges n is divisible by 2 then you can eat n / 2 oranges. If the number of remaining oranges n is divisible by 3 then you can eat 2 * (n / 3) oranges. You can only choose one of the actions per day. Given the integer n, return the minimum number of days to eat n oranges.

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

Example 2

Step 02

Core Insight

What unlocks the optimal approach

In each step, choose between 2 options: minOranges = 1 + min( (n%2) + f(n/2), (n%3) + f(n/3) ) where f(n) is the minimum number of days to eat n oranges.

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

Largest constraint values

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 #1553: Minimum Number of Days to Eat N Oranges
class Solution {
    private Map<Integer, Integer> f = new HashMap<>();

    public int minDays(int n) {
        return dfs(n);
    }

    private int dfs(int n) {
        if (n < 2) {
            return n;
        }
        if (f.containsKey(n)) {
            return f.get(n);
        }
        int res = 1 + Math.min(n % 2 + dfs(n / 2), n % 3 + dfs(n / 3));
        f.put(n, res);
        return res;
    }
}

// Accepted solution for LeetCode #1553: Minimum Number of Days to Eat N Oranges
func minDays(n int) int {
	f := map[int]int{0: 0, 1: 1}
	var dfs func(int) int
	dfs = func(n int) int {
		if v, ok := f[n]; ok {
			return v
		}
		res := 1 + min(n%2+dfs(n/2), n%3+dfs(n/3))
		f[n] = res
		return res
	}
	return dfs(n)
}

# Accepted solution for LeetCode #1553: Minimum Number of Days to Eat N Oranges
class Solution:
    def minDays(self, n: int) -> int:
        @cache
        def dfs(n: int) -> int:
            if n < 2:
                return n
            return 1 + min(n % 2 + dfs(n // 2), n % 3 + dfs(n // 3))

        return dfs(n)

// Accepted solution for LeetCode #1553: Minimum Number of Days to Eat N Oranges
struct Solution;

use std::collections::HashMap;

impl Solution {
    fn min_days(n: i32) -> i32 {
        let mut memo: HashMap<i32, i32> = HashMap::new();
        Self::dp(n, &mut memo)
    }

    fn dp(n: i32, memo: &mut HashMap<i32, i32>) -> i32 {
        if n <= 1 {
            n
        } else {
            if let Some(&res) = memo.get(&n) {
                return res;
            }
            let res = 1 + (n % 2 + Self::dp(n / 2, memo)).min(n % 3 + Self::dp(n / 3, memo));
            memo.insert(n, res);
            res
        }
    }
}

#[test]
fn test() {
    let n = 10;
    let res = 4;
    assert_eq!(Solution::min_days(n), res);
    let n = 6;
    let res = 3;
    assert_eq!(Solution::min_days(n), res);
    let n = 1;
    let res = 1;
    assert_eq!(Solution::min_days(n), res);
    let n = 56;
    let res = 6;
    assert_eq!(Solution::min_days(n), res);
    let n = 9209408;
    let res = 23;
    assert_eq!(Solution::min_days(n), res);
    let n = 84806671;
    let res = 32;
    assert_eq!(Solution::min_days(n), res);
}

// Accepted solution for LeetCode #1553: Minimum Number of Days to Eat N Oranges
function minDays(n: number): number {
    const f: Record<number, number> = {};
    const dfs = (n: number): number => {
        if (n < 2) {
            return n;
        }
        if (f[n] !== undefined) {
            return f[n];
        }
        f[n] = 1 + Math.min((n % 2) + dfs((n / 2) | 0), (n % 3) + dfs((n / 3) | 0));
        return f[n];
    };
    return dfs(n);
}

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(log^2 n)

Space

O(log^2 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.

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.