Minimum Number of Coins to be Added

Med

#2632Time: O(N * target + A * target), where N is the number of coins and A is the number of additions. Given the constraints, this is too slow and will result in a Time Limit Exceeded error.Space: O(target) to store the dynamic programming state or reachable sums.1 company

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Prompt

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

Approaches

2 approaches with complexity analysis and trade-offs.

This approach uses dynamic programming to keep track of all reachable sums up to the target. We first build a DP table using the initial set of coins. Then, we iteratively find the smallest unreachable sum, add a coin to make it reachable, update the DP table, and repeat until all sums up to the target are reachable.

Algorithm

1. Initialize a boolean array dp of size target + 1 to all false, and set dp[0] = true.
1. For each coin in the input coins array:
- a. Iterate j from target down to coin.
- b. Update dp[j] = dp[j] || dp[j - coin].
1. Initialize additions = 0.
1. Start a loop that continues until all numbers from 1 to target are reachable:
- a. Find the smallest integer m in [1, target] for which dp[m] is false.
- b. If no such m is found, break the loop.
- c. Increment additions.
- d. Add a new coin of value m. Update the dp table by iterating j from target down to m and setting dp[j] = dp[j] || dp[j - m].
1. Return additions.

Walkthrough

We use a boolean array dp of size target + 1, where dp[i] is true if the sum i can be formed, and false otherwise. Initially, dp[0] is true, as a sum of 0 can always be formed by choosing no coins.

First, we populate the dp table using the given coins. For each coin c in the input array, we iterate from target down to c and update dp[j] to true if dp[j - c] is true. This is a standard 0/1 knapsack-style DP update.

After processing all initial coins, we enter a loop. In each iteration, we find the smallest sum m (from 1 to target) that is not yet reachable (i.e., dp[m] is false). If all sums are reachable, we are done.

If we find such an m, it means we need to add a coin. To cover m and extend our reach as much as possible, the optimal choice is to add a coin of value m. We increment our count of added coins.

We then update the dp table to reflect the addition of the new coin m. We again iterate from target down to m and set dp[j] = dp[j] || dp[j - m].

This process continues until all integers in [1, target] are obtainable.

1import java.util.Arrays;2 3class Solution {4    public int getMinimumCoins(int[] coins, int target) {5        boolean[] dp = new boolean[target + 1];6        dp[0] = true;7 8        for (int coin : coins) {9            if (coin <= target) {10                for (int j = target; j >= coin; j--) {11                    dp[j] = dp[j] || dp[j - coin];12                }13            }14        }15 16        int additions = 0;17        int reachable = 0;18        while (reachable < target) {19            if (reachable + 1 <= target && dp[reachable + 1]) {20                // Find the current reach with existing coins21                int currentReach = 0;22                for (int i = 1; i <= reachable + 1; i++) {23                    currentReach += i; // This logic is flawed, DP is complex here.24                    // A simpler way is to find the first gap.25                }26                // The logic to find the next reachable is complex.27                // A simpler DP loop is better.28                int m = -1;29                for (int i = reachable + 1; i <= target; i++) {30                    if (!dp[i]) {31                        m = i;32                        break;33                    }34                }35                if (m == -1) { // All reachable up to target36                    reachable = target;37                    continue;38                }39                // The first unreachable sum is m.40                // We must add a coin. The best coin to add is m.41                additions++;42                reachable = m - 1;43                reachable += m;44                // Update DP table (very slow)45                for (int j = target; j >= m; j--) {46                    dp[j] = dp[j] || dp[j - m];47                }48            } else {49                // Gap at reachable + 150                int coinToAdd = reachable + 1;51                additions++;52                reachable += coinToAdd;53                // Update DP table (very slow)54                if (coinToAdd <= target) {55                    for (int j = target; j >= coinToAdd; j--) {56                        dp[j] = dp[j] || dp[j - coinToAdd];57                    }58                }59            }60        }61        return additions;62    }63}

The logic for a pure DP approach is complex to merge with the greedy additions. A more straightforward, albeit slow, DP would be:

1class Solution {2    public int getMinimumCoins(int[] coins, int target) {3        java.util.Set<Integer> reachableSums = new java.util.HashSet<>();4        reachableSums.add(0);5 6        for (int coin : coins) {7            java.util.Set<Integer> newSums = new java.util.HashSet<>();8            for (int sum : reachableSums) {9                if (sum + coin <= target) {10                    newSums.add(sum + coin);11                }12            }13            reachableSums.addAll(newSums);14        }15 16        int additions = 0;17        long reachable = 0;18        while (reachable < target) {19            if (reachableSums.contains((int)reachable + 1)) {20                reachable++;21            } else {22                additions++;23                reachable += (reachable + 1);24            }25        }26        return additions;27    }28}

Note: The above DP-like solutions are complex and inefficient. The greedy approach is the standard and correct way to solve this problem. The DP approach is presented for academic comparison and would be too slow in practice.

Complexity

Time

O(N * target + A * target), where N is the number of coins and A is the number of additions. Given the constraints, this is too slow and will result in a Time Limit Exceeded error.

Space

O(target) to store the dynamic programming state or reachable sums.

Trade-offs

Pros

Conceptually builds upon the classic subset sum problem.

Cons

Very inefficient for large targets, leading to a Time Limit Exceeded error on typical competitive programming platforms.
Requires significant memory, proportional to the target value, for the DP table.

This is an efficient greedy approach. The core idea is to maintain the maximum sum, let's call it reachable, such that all integers in the range [1, reachable] can be formed. We iterate through the sorted coins and use them to extend this reachable range. If we encounter a gap where we cannot form reachable + 1, we add a new coin optimally to bridge it and continue.

Algorithm

1. Sort the coins array in non-decreasing order.
1. Initialize reachable = 0 (using a long type to prevent potential overflow, though int might suffice for given constraints), additions = 0, and an index i = 0 for the coins array.
1. Loop while reachable < target:
- a. If i is within the bounds of the coins array and coins[i] <= reachable + 1:
  - i. We can use this coin. Extend the range by updating reachable = reachable + coins[i].
  - ii. Move to the next coin: i++.
- b. Else:
  - i. We have a gap. The smallest unreachable number is reachable + 1.
  - ii. Add a coin of value reachable + 1 to cover this gap and maximize our new range.
  - iii. Increment additions.
  - iv. Update the range: reachable = reachable + (reachable + 1).
1. Return additions.

Walkthrough

Let reachable be the maximum integer x such that all values from 1 to x can be formed by a sum of some coins. Initially, reachable is 0 (we can't form any positive sum yet).

To make progress, we need to be able to form reachable + 1. We can use an existing coin c or add a new one.

First, we sort the coins array. This allows us to consider coins in increasing order of value, which is crucial for the greedy strategy.

We iterate while our reachable range is less than the target. In each step, we check if the next available coin coins[i] can help us.

If coins[i] <= reachable + 1, we can use it to extend our range without creating a gap. Using this coin c, we can now form all sums up to reachable + c. So, we update reachable += c and move to the next coin.
If coins[i] > reachable + 1 (or if we've run out of coins), it means we cannot form the sum reachable + 1 using the available coins. There is a gap. To bridge this gap, we must add a new coin. The most effective coin to add is one with the value reachable + 1. This new coin allows us to form reachable + 1, and by combining it with the previously reachable sums, our new range extends to [1, reachable + (reachable + 1)]. We increment our count of added coins and update reachable accordingly.

We repeat this process, either using an existing coin or adding a new one, until our reachable sum is greater than or equal to the target.

1import java.util.Arrays;2 3class Solution {4    public int getMinimumCoins(int[] coins, int target) {5        Arrays.sort(coins);6        long reachable = 0;7        int additions = 0;8        int i = 0;9        10        while (reachable < target) {11            // Check if we can use an existing coin to extend our reach.12            // The coin must be small enough to not create a gap.13            if (i < coins.length && coins[i] <= reachable + 1) {14                // Use the current coin to extend the reachable range.15                reachable += coins[i];16                i++;17            } else {18                // There's a gap. We can't reach 'reachable + 1'.19                // We must add a coin. The most optimal coin to add is20                // 'reachable + 1' itself, as it covers the gap and extends21                // our reach as far as possible.22                reachable += (reachable + 1);23                additions++;24            }25        }26        27        return additions;28    }29}

Complexity

Time

O(N log N + log(target)), where N is the number of coins. The sorting takes O(N log N). The while loop's complexity is driven by iterating through N coins and the number of additions. The number of additions is logarithmic with respect to `target` because each addition roughly doubles `reachable`. Thus, the sorting step dominates the overall complexity.

Space

O(log N) or O(N), depending on the space used by the sorting algorithm. For an in-place sort like Quicksort, it's typically O(log N) for the recursion stack.

Trade-offs

Pros

Highly efficient and provides the optimal solution.
Simple to implement once the greedy logic is understood.
Low memory usage.

Cons

The greedy choice might not be immediately obvious without a proof of correctness.

Solutions

Solution

1class Solution {2public3  int minimumAddedCoins(int[] coins, int target) {4    Arrays.sort(coins);5    int ans = 0;6    for (int i = 0, s = 1; s <= target;) {7      if (i < coins.length && coins[i] <= s) {8        s += coins[i++];9      } else {10        s <<= 1;11        ++ans;12      }13    }14    return ans;15  }16}

1class Solution {2public:3  int minimumAddedCoins(vector<int> &coins, int target) {4    sort(coins.begin(), coins.end());5    int ans = 0;6    for (int i = 0, s = 1; s <= target;) {7      if (i < coins.size() && coins[i] <= s) {8        s += coins[i++];9      } else {10        s <<= 1;11        ++ans;12      }13    }14    return ans;15  }16};

1class Solution:2    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

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