LeetCode #2952 — MEDIUM

Minimum Number of Coins to be Added

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

The Problem

Problem Statement

You are given a 0-indexed integer array coins, representing the values of the coins available, and an integer target.

An integer x is obtainable if there exists a subsequence of coins that sums to x.

Return the minimum number of coins of any value that need to be added to the array so that every integer in the range [1, target] is obtainable.

A subsequence of an array is a new non-empty array that is formed from the original array by deleting some (possibly none) of the elements without disturbing the relative positions of the remaining elements.

Example 1:

Input: coins = [1,4,10], target = 19
Output: 2
Explanation: We need to add coins 2 and 8. The resulting array will be [1,2,4,8,10].
It can be shown that all integers from 1 to 19 are obtainable from the resulting array, and that 2 is the minimum number of coins that need to be added to the array.

Example 2:

Input: coins = [1,4,10,5,7,19], target = 19
Output: 1
Explanation: We only need to add the coin 2. The resulting array will be [1,2,4,5,7,10,19].
It can be shown that all integers from 1 to 19 are obtainable from the resulting array, and that 1 is the minimum number of coins that need to be added to the array.

Example 3:

Input: coins = [1,1,1], target = 20
Output: 3
Explanation: We need to add coins 4, 8, and 16. The resulting array will be [1,1,1,4,8,16].
It can be shown that all integers from 1 to 20 are obtainable from the resulting array, and that 3 is the minimum number of coins that need to be added to the array.

Constraints:

1 <= target <= 10⁵
1 <= coins.length <= 10⁵
1 <= coins[i] <= target

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array coins, representing the values of the coins available, and an integer target. An integer x is obtainable if there exists a subsequence of coins that sums to x. Return the minimum number of coins of any value that need to be added to the array so that every integer in the range [1, target] is obtainable. A subsequence of an array is a new non-empty array that is formed from the original array by deleting some (possibly none) of the elements without disturbing the relative positions of the remaining elements.

Baseline thinking

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

Pattern signal: Array · Greedy

Example 1

[1,4,10]
19

Example 2

[1,4,10,5,7,19]
19

Example 3

[1,1,1]
20

Core Insight

What unlocks the optimal approach

Sort the coins array and maintain the smallest sum that is unobtainable by induction.
If we don’t use any coins, the smallest integer that we cannot obtain by sum is <code>1</code>. Suppose currently, for a fixed set of the first several coins the smallest integer that we cannot obtain is <code>x + 1</code>, namely we can form all integers in the range <code>[1, x]</code> but not <code>x + 1</code>.
If the next unused coin’s value is NOT <code>x + 1</code> (note the array is sorted), we have to add <code>x + 1</code> to the array. After this addition, we can form all values from <code>x + 1</code> to <code>2 * x + 1</code> by adding <code>x + 1</code> in <code>[1, x]</code>'s formations. So now we can form all the numbers of <code>[1, 2 * x + 1]</code>. After this iteration the new value of <code>x</code> becomes <code>2 * x + 1</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 #2952: Minimum Number of Coins to be Added
class Solution {
    public int minimumAddedCoins(int[] coins, int target) {
        Arrays.sort(coins);
        int ans = 0;
        for (int i = 0, s = 1; s <= target;) {
            if (i < coins.length && coins[i] <= s) {
                s += coins[i++];
            } else {
                s <<= 1;
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2952: Minimum Number of Coins to be Added
func minimumAddedCoins(coins []int, target int) (ans int) {
	slices.Sort(coins)
	for i, s := 0, 1; s <= target; {
		if i < len(coins) && coins[i] <= s {
			s += coins[i]
			i++
		} else {
			s <<= 1
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #2952: Minimum Number of Coins to be Added
class Solution:
    def minimumAddedCoins(self, coins: List[int], target: int) -> int:
        coins.sort()
        s = 1
        ans = i = 0
        while s <= target:
            if i < len(coins) and coins[i] <= s:
                s += coins[i]
                i += 1
            else:
                s <<= 1
                ans += 1
        return ans

// Accepted solution for LeetCode #2952: Minimum Number of Coins to be Added
/**
 * [2952] Minimum Number of Coins to be Added
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn minimum_added_coins(coins: Vec<i32>, target: i32) -> i32 {
        let mut coins = coins;
        coins.sort_unstable();

        let mut result = 0;
        let mut i = 0;
        let mut x = 1;

        while x <= target {
            if i < coins.len() && coins[i] <= x {
                x += coins[i];
                i += 1;
            } else {
                x = x * 2;
                result += 1;
            }
        }

        result
    }
}

// submission codes end

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

    #[test]
    fn test_2952() {
        assert_eq!(2, Solution::minimum_added_coins(vec![1, 4, 10], 19));
        assert_eq!(
            0,
            Solution::minimum_added_coins(vec![1, 2, 4, 6, 7, 9, 9, 10], 48)
        );
    }
}

// Accepted solution for LeetCode #2952: Minimum Number of Coins to be Added
function minimumAddedCoins(coins: number[], target: number): number {
    coins.sort((a, b) => a - b);
    let ans = 0;
    for (let i = 0, s = 1; s <= target; ) {
        if (i < coins.length && coins[i] <= s) {
            s += coins[i++];
        } else {
            s <<= 1;
            ++ans;
        }
    }
    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 × log n)

Space

O(log n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.