Print Right Diamond Number Pattern Type9

June 23, 2019

8 comments on “Print Right Diamond Number Pattern Type9”


  • Ajitha

    n=4
    start1=2
    for i in range(1,n+1):
    num1=start1
    for j in range(i):
    print(num1,end=””)
    num1-=1
    print()
    start1=(start1+i)+1
    start2=11
    for i in range(n,0,-1):
    num2=start2
    for j in range(i):
    print(num2,end=””)
    num2-=1
    print()
    start2=start2-i


  • shraddhagpatil7

    Solution in C++

    Code:
    #include
    using namespace std;
    int main() {

    int numm;
    int num=2;
    for(int i=0;i<=3;i++){
    num=num+i;
    numm=num+i;
    for(int j=0;j<=i;j++){

    cout<<numm<<" ";
    numm–;
    }
    cout<=0;i–){

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

    cout<<num2<<" ";
    num2–;
    }
    cout<<endl;

    }

    return 0;
    }


  • abhishekgupta7554

    void threeStarPattern(){
    int n; cin>>n;
    int cnt=2;
    for(int i=1;i<=n;i++){
    string temp="";
    for(int j=1;j<=i;j++){
    temp+=to_string(cnt);
    cnt++;
    }
    reverse(temp.begin(),temp.end());
    cout<<temp;
    cout<<endl;
    }
    int val=((n)*(n+1))/2 +1;
    for(int i=1;i<=n;i++){
    stackst;
    for(int j=1;j<=n-i+1;j++){
    st.push(val);
    val–;
    }
    string temp="";
    while(!st.empty()){
    int top=st.top();
    temp+= to_string(top);
    st.pop();
    }
    reverse(temp.begin(),temp.end());
    cout<<temp;
    cout<<endl;
    }
    }


  • varikutisrinivas123

    n=5
    k=3
    for i in range(1,n+1):
    z=k
    for j in range(i):
    print(z,end=””)
    z-=1
    print()
    if i!=n:
    k=k+(i+1)
    for i in range(n,0,-1):
    z=k
    for j in range(i):
    print(z,end=””)
    z-=1
    print()
    k=k-i