max no of ways (coin change) (1,2 & 2,1 counted once)

// unbounded knapsack // only difference from 01 is that we can choose one element multiple times.
// here value is weight and as we have to count the number of ways adding value doesnt matter
//knapsack is total sum
int solve(vector<int> value, int knapsack){
    int  n = value.size();
    vector<vector<int>> dp (n+1 ,vecotr<int>(knapsack+1,0));
    // a represents the ath element .. b represents the knapsack present
    for(int a=0;a<=n;a++) dp[a][0] = 1; // dont select any coin
    for(int a=1;a<=n;a++){
        for(int b=1;b<=knapsack;b++){
            if(value[a-1]<=b){
            dp[a][b] = dp[a][b-value[a-1]] + dp[a-1][b]; //here we can choose ith element again
            }else{
            dp[a][b] = dp[a-1][b]; // cant choose the ath element becuase its weight is more
            }
        }
    }
    return dp[n][knapsack];

}