Coin Change

Given a value N, find the number of ways to make change for N cents, if we have infinite supply of each of S = { S1, S2, .. , SM } valued coins.


Example 1:

Input:
n = 4 , m = 3
S[] = {1,2,3}
Output: 4
Explanation: Four Possible ways are:
{1,1,1,1},{1,1,2},{2,2},{1,3}.
Example 2:

Input:
n = 10 , m = 4
S[] ={2,5,3,6}
Output: 5
Explanation: Five Possible ways are:
{2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5}
and {5,5}.

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class GFG {


static int xorOfArray(int arr[], int n)
{

int xor_arr = 0;


for (int i = 0; i < n; i++) {


xor_arr = xor_arr ^ arr[i];
}


return xor_arr;
}


public static void main (String[] args)
{

int arr[] = { 3, 9, 12, 13, 15 };
int n = arr.length;


System.out.println(xorOfArray(arr, n));

}
}

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#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
char B[105];
int main(){
int n;
int ans=0;
scanf("%d", &n);
scanf("%s", B);
for(int i=2; B[i]; i++){
if(B[i-2] == '0' && B[i-1] == '1' && B[i] == '0') B[i] = '1', ans++;
}
printf("%d", ans);
return 0;
}

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ArrayList<Integer> ar = new ArrayList<Integer>();
ArrayList<Integer> ar1 = new ArrayList<Integer>();
ArrayList<Integer> arr2= new ArrayList<Integer>();
ArrayList<Integer> sza= new ArrayList<Integer>();
ArrayList<Integer> p_n= new ArrayList<Integer>();

Scanner sc = new Scanner(System.in);

int a_size= sc.nextInt();
int m_size=sc.nextInt();
int[] a = new int[a_size];
for(int i=0;i<a_size;i++){
a[i]=sc.nextInt();
arr2.add(a[i]);
}
int[] p = new int[m_size];
for(int i=0;i<m_size;i++){
p[i]=sc.nextInt();
p_n.add(p[i]);
}
int[] sz=new int[m_size];
for(int i=0;i<m_size;i++){
sz[i]=sc.nextInt();
sza.add(sz[i]);
}

for(int i=0;i<a_size;i++)
{
for(int j=0;j<m_size;j++)
{
if(a[i]==p[j])
{
ar.add(p[j]);
}
}
}
int [] fr = new int [ar.size()];
int visited = -1;
for(int i = 0; i < ar.size(); i++){
int count = 1;
for(int j = i+1; j < ar.size(); j++){
if(ar.get(i).equals(ar.get(j))){
count++;
fr[j] = visited;
}
}
if(fr[i] != visited)
fr[i] = count;
}

for(int i = 0; i < fr.length; i++){
if(fr[i] != visited)
ar1.add(fr[i]);
}
ArrayList<Integer> pnew = new ArrayList<Integer>();
for(int l=0;l<sz.length;l++)
{
for(int h=0;h<sza.get(l);h++)
{
pnew.add(p_n.get(l));
}
}
int count=0;
for(int i = 0; i < pnew.size(); i++)
{
for(int j = i+1; j < pnew.size(); j++)
{
if(pnew.get(i).equals(pnew.get(j)))
{
count++;
}
}
}
System.out.println(count);
}
}


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