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 :



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                        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


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