**Split Digits to Sum Closest To Target - Google Top Interview Questions**

### Problem Statement :

You are given a string s and an integer target. s represents a decimal number containing digits from 0 to 9. You can partition s into as many parts as you want and take the sum of its parts. Afterwards, return the minimum possible absolute difference to target. Constraints 1 ≤ n, target ≤ 1,000 where n is the length of s Example 1 Input s = "112" target = 10 Output 3 Explanation We can partition s into "1" + "12" which sums to 13 and abs(13 - 10) = 3. Example 2 Input s = "500" target = 300 Output 200 Explanation The best we can do is split s into just "500" and abs(300 - 500) = 200 Example 3 Input s = "1" target = 9 Output 8 Explanation We can only split s into just "1".

### Solution :

` ````
Solution in C++ :
int help(string& s, int target, int i, vector<vector<int>>& dp) {
if (target < 0) {
int ret = 0;
for (int k = i; k < s.length(); k++) {
ret += (s[k] - '0');
}
return ret - target;
}
if (i == s.length()) return target;
if (dp[i][target] != -1) return dp[i][target];
int ret = 1e6, curr = 0;
for (int t = i; t < i + 4 and t < s.length(); t++) {
curr = curr * 10 + s[t] - '0';
ret = min(ret, help(s, target - curr, t + 1, dp));
}
return dp[i][target] = ret;
}
int solve(string s, int target) {
int n = s.length();
vector<vector<int>> dp(n, vector<int>(target + 1, -1));
return help(s, target, 0, dp);
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(String s, int target) {
Set<Integer>[] sets = new HashSet[s.length() + 1];
for (int i = 0; i <= s.length(); i++) sets[i] = new HashSet<>();
sets[0].add(0);
for (int i = 0, i2 = 1; i != s.length(); i++, i2++) { // i2 is my index
Set<Integer> seti = sets[i2];
for (int j = 1; j <= 4; j++) {
int start = i + 1 - j;
if (start < 0)
break;
final int val = Integer.parseInt(s.substring(start, i + 1));
for (Integer prev : sets[start]) {
int nextval = prev + val;
if (nextval <= 2000)
seti.add(nextval);
}
}
}
int res = Integer.MAX_VALUE;
for (int v : sets[s.length()]) res = Math.min(res, Math.abs(target - v));
return res;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s, target):
n = len(s)
s = [int(c) for c in s]
# For each index, precompute the index of the next nonzero digit.
last = n
next_nonzero = [0] * n
for i in reversed(range(n)):
if s[i] != 0:
last = i
next_nonzero[i] = last
# Pick an arbitrary "bad" solution. We won't consider any solution worse than this one.
baseline = abs(target - sum(s))
@cache
def dp(i, t):
if i == n:
# base case
return abs(t)
if s[i] == 0:
# Skip leading zeroes, as they always contribute nothing.
return dp(next_nonzero[i], t)
ans = baseline
cur = 0
for j in range(i, n):
cur = cur * 10 + s[j]
if cur >= t + baseline:
break
if j == n - 1 or cur != 0:
ans = min(ans, dp(j + 1, t - cur))
return ans
return dp(0, target)
```

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