Old Top Coder riddle: Making a number by inserting +

Question

I am thinking about this topcoder problem.

Given a string of digits, find the minimum number of additions required for the string to equal some target n

Answer 1

I haven't given it much thought yet, but if you scroll down you can see a link to the contest it was from, and from there you can see the solvers' solutions. Here's one in C#.

using System; 
using System.Text; 
using System.Text.RegularExpressions; 
using System.Collections; 

public class QuickSums { 
    public int minSums(string numbers, int sum) { 
        int[] arr = new int[numbers.Length]; 
    for (int i = 0 ; i < arr.Length; i++)       
      arr[i] = 0; 

    int min = 15; 

    while (arr[arr.Length - 1] != 2)     
    { 
      arr[0]++; 
      for (int i = 0; i < arr.Length - 1; i++) 
        if (arr[i] == 2)  
        { 
          arr[i] = 0; 
          arr[i + 1]++; 
        } 

      String newString = ""; 
      for (int i = 0; i < numbers.Length; i++) 
      { 
        newString+=numbers[i]; 
        if (arr[i] == 1) 
          newString+="+"; 
      } 

      String[] nums = newString.Split('+'); 
      int sum1 = 0; 
      for (int i = 0; i < nums.Length; i++) 
        try  
        { 
          sum1 += Int32.Parse(nums[i]); 
        } 
        catch 
        { 
        } 

      if (sum == sum1 && nums.Length - 1 < min) 
        min = nums.Length - 1; 
    } 

    if (min == 15) 
      return -1; 

    return min; 
  } 

} 

Answer 2

Seems to be too late .. but just read some comments and answers here which say no to dp approach . But it is a very straightforward dp similar to rod-cutting problem:

To get the essence:

int val[N][N];
int dp[N][T];

val[i][j]: numerical value of s[i..j] including both i and j

val[i][j] can be easily computed using dynamic programming approach in O(N^2) time

dp[i][j] : Minimum no of '+' symbols to be inserted in s[0..i] to get the required sum j

dp[i][j] = min( 1+dp[k][j-val[k+1][j]] ) over all k such that 0<=k<=i and val[k][j]>0

In simple terms , to compute dp[i][j] you assume the position k of last '+' symbol and then recur for s[0..k]

Answer 3

dynamic programming :

public class QuickSums {

public static int req(int n, int[] digits, int sum) {

    if (n == 0) {
        if (sum == 0)
            return 0;
        else
            return -1;
    } else if (n == 1) {
        if (sum == digits[0]) {
            return 0;
        } else {
            return -1;
        }
    }

    int deg = 1;
    int red = 0;

    int opt = 100000;
    int split = -1;

    for (int i=0; i<n;i++) {
        red += digits[n-i-1] * deg;

        int t = req(n-i-1,digits,sum - red);
        if (t != -1 && t <= opt) {
            opt = t;
            split = i;
        }
        deg = deg*10;
    }

    if (opt == 100000) 
        return -1;
    if (split == n-1) 
        return opt;
    else
        return opt + 1;

}

public static int solve (String digits,int sum) {
   int [] dig = new int[digits.length()];
   for (int i=0;i<digits.length();i++) {
       dig[i] = digits.charAt(i) - 48;
   }

    return req(digits.length(), dig, sum);
}


public static void doit() {
    String digits = "9230560001";
    int sum = 71;

    int result = solve(digits, sum);
    System.out.println(result);
}

Answer 4

I think the problem is similar to Matrix Chain Multiplication problem where we have to put braces for least multiplication. Here braces represent '+'. So I think it could be solved by similar dp approach.. Will try to implement it.

Answer 5

Because input length is small (10) all possible ways (which can be found by a simple binary counter of length 10) is small (2^10 = 1024), so your algorithm is fast enough and returns valid result, and IMO there is no need to improve it.

In all until your solution works fine in time and memory and other given constrains, there is no need to do micro optimization. e.g this case as akappa offered can be solved with DP like DP in two-Partition problem, but when your algorithm is fast there is no need to do this and may be adding some big constant or making code unreadable.

I just offer parse digits of string one time (in array of length 10) to prevent from too many string parsing, and just use a*10^k + ... (Also you can calculate 10^k for k=0..9 in startup and save its value).

Answer 6

It's certainly not optimal. If, for example, you are given the string "1234567890" and the target is a three-digit number, you know that you have to split the string into at least four parts, so you need not check 0, 1, or 2 inserts. Also, the target limits the range of admissible insertion positions. Both points have small impact for short strings, but can make a huge difference for longer ones. However, I suspect there's a vastly better method, smells a bit of DP.