# 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<>();
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)
}
}
}
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)```
```

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

## Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

## Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

## Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -