# 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 =  * 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)```
```

## Dynamic Array

Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing. Create an integer, lastAnswer, and initialize it to 0. There are 2 types of queries that can be performed on the list of sequences: 1. Query: 1 x y a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.

## Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

## Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

## Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

## Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink