Degree of Polynomial Solution

Problem

In mathematics, the degree of polynomials in one variable is the highest power of the variable in the algebraic expression with non-zero coefficient.

Chef has a polynomial in one variable $x$ with $N$ terms. The polynomial looks like $A_0\cdot x^0 + A_1\cdot x^1 + \ldots + A_{N-2}\cdot x^{N-2} + A_{N-1}\cdot x^{N-1}$ where $A_{i-1}$ denotes the coefficient of the $i^{th}$ term $x^{i-1}$ for all $(1\le i\le N)$.

Find the degree of the polynomial.

Note: It is guaranteed that there exists at least one term with non-zero coefficient.

Input Format

• First line will contain $T$, number of test cases. Then the test cases follow.
• First line of each test case contains of a single integer $N$ - the number of terms in the polynomial.
• Second line of each test case contains of $N$ space-separated integers - the $i^{th}$ integer $A_{i-1}$ corresponds to the coefficient of $x^{i-1}$.

Output Format

For each test case, output in a single line, the degree of the polynomial.

Constraints

• $1 \leq T \leq 100$
• $1 \leq N \leq 1000$
• $-1000 \le A_i \le 1000$
• $A_i \ne 0$ for at least one $(0\le i \lt N)$.

Sample 1:

Input

Output

4
1
5
2
-3 3
3
0 0 5
4
1 2 4 0

0
1
2
2

Explanation:

Test case $1$: There is only one term $x^0$ with coefficient $5$. Thus, we are given a constant polynomial and the degree is $0$.

Test case $2$: The polynomial is $-3\cdot x^0 + 3\cdot x^1 = -3 + 3\cdot x$. Thus, the highest power of $x$ with non-zero coefficient is $1$.

Test case $3$: The polynomial is $0\cdot x^0 + 0\cdot x^1 + 5\cdot x^2= 0+0 + 5\cdot x^2$. Thus, the highest power of $x$ with non-zero coefficient is $2$.

Test case $4$: The polynomial is $1\cdot x^0 + 2\cdot x^1+ 4\cdot x^2 + 0\cdot x^3= 1 + 2\cdot x + 4\cdot x^2$. Thus, the highest power of $x$ with non-zero coefficient is $2$.

Program :

 import java.util.*;

import java.lang.Math.*;

class Codechef

{

    public static void main(String args[])

    {

        Scanner es = new Scanner(System.in);

        int t,n,i,j,p,k;

        int a[]= new int[1000];

        t=es.nextInt();

        for(i=0;i<t;i++)

        {

    n=es.nextInt();

            p=n-1;

            for(j=0;j<n;j++)

                a[j]=es.nextInt();

            for(k=0;k<n;k++)

            {

                if(a[k]!=0)

                p=k;

            }

            System.out.println(p);

        }

        es.close();

    }

}