# Number of Operations to Decrement Target to Zero - Google Top Interview Questions

### Problem Statement :

```You are given a list of positive integers nums and an integer target.

Consider an operation where we remove a number v from either the front or the back of nums and decrement target by v.

Return the minimum number of operations required to decrement target to zero. If it's not possible, return -1.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [3, 1, 1, 2, 5, 1, 1]

target = 7

Output

3

Explanation

We can remove 1, 1 and 5 from the back to decrement target to zero.

Example 2

Input

nums = [2, 4]

target = 7

Output

-1

Explanation

There's no way to decrement target = 7 to zero.```

### Solution :

```                        ```Solution in C++ :

int solve(vector<int>& numbers, int target) {
vector<int> prefix_sums = {0};
partial_sum(numbers.begin(), numbers.end(), back_inserter(prefix_sums));

int length = numbers.size();
int shortest = length + 1;

for (int left = 0; left <= length; ++left) {
int target_for_right = prefix_sums[length] + prefix_sums[left] - target;
auto it = lower_bound(prefix_sums.begin(), prefix_sums.end(), target_for_right);
int right = distance(prefix_sums.begin(), it);
bool match = prefix_sums[right] == target_for_right;

if (match) {
shortest = min(shortest, length - right + left);
}
}

return shortest == length + 1 ? -1 : shortest;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int solve(int[] nums, int target) {
int n = nums.length;
if (n == 0) {
if (target == 0)
return 0;
else
return -1;
}
if (target == 0)
return 0;
int sum = 0;
for (int num : nums) sum += num;
if (sum == target)
return n;
target = sum - target;
HashMap<Integer, Integer> map = new HashMap<>();
int t = 0;
int ans = Integer.MAX_VALUE;
for (int i = 0; i < n; i++) {
t += nums[i];
if (t == target)
ans = Math.min(n - 1 - i, ans);
if (map.containsKey(t - target))
ans = Math.min(ans, n - (i - map.get(t - target)));
map.put(t, i);
}
return ans == Integer.MAX_VALUE ? -1 : ans;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, nums, target):
if target == 0:
return 0
psum = sum(nums)
if psum < target:
return -1
ans = len(nums) + 1
ssum = 0
suffixcnt = 0
for i in range(len(nums) - 1, -1, -1):
psum -= nums[i]
while psum + ssum < target:
ssum += nums[-1 - suffixcnt]
suffixcnt += 1
if psum + ssum == target:
ans = min(ans, i + suffixcnt)
if ans > len(nums):
ans = -1
return ans```
```

## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio

## Cube Summation

You are given a 3-D Matrix in which each block contains 0 initially. The first block is defined by the coordinate (1,1,1) and the last block is defined by the coordinate (N,N,N). There are two types of queries. UPDATE x y z W updates the value of block (x,y,z) to W. QUERY x1 y1 z1 x2 y2 z2 calculates the sum of the value of blocks whose x coordinate is between x1 and x2 (inclusive), y coor

## Direct Connections

Enter-View ( EV ) is a linear, street-like country. By linear, we mean all the cities of the country are placed on a single straight line - the x -axis. Thus every city's position can be defined by a single coordinate, xi, the distance from the left borderline of the country. You can treat all cities as single points. Unfortunately, the dictator of telecommunication of EV (Mr. S. Treat Jr.) do

## Subsequence Weighting

A subsequence of a sequence is a sequence which is obtained by deleting zero or more elements from the sequence. You are given a sequence A in which every element is a pair of integers i.e A = [(a1, w1), (a2, w2),..., (aN, wN)]. For a subseqence B = [(b1, v1), (b2, v2), ...., (bM, vM)] of the given sequence : We call it increasing if for every i (1 <= i < M ) , bi < bi+1. Weight(B) =