equal sum partition

// check whether the given equal sum partition is possible or not.
// here we dont have wt array
// s1 +s2 = total sum
// as s1 = s2 ====> check whether total sum/2 exists or not.

int solve(vector<int> value){
    int  n = value.size();
    int sum=0;
    for(int a=0;a<n;a++)sum+=value[a];
    if(sum%2 == 1) return 0; // as odd sum can't be partitioned into two .
    vector<vector<bool>> dp (n+1 ,vector<bool>(sum+1,false));
    // a represents the ath element .. b represents the knapsack present

    for(int a=0;a<=n;a++)dp[a][0] = true;// sum 0 is possible even if do not select anything

    for(int a=1;a<=n;a++){
        for(int b=1;b<=sum;b++){
            if(value[a-1]<=b){
            dp[a][b] = dp[a-1][b-value[a-1]] || dp[a-1][b]; //either i choose the ath element or not
            }else{
            dp[a][b] = dp[a-1][b]; // cant choose the ath element becuase its weight is more
            }
        }
    }
    return dp[n][sum/2];

}