Programming Problems and Solutions in Java


Problem #1


Write a program in Java which accept 2 integer start & repeat and print the pattern such that starting and ending number of pattern is - start and total numbers or rows in the pattern is equal to (repeat*2). Number in the pattern should be increase by 1 in each rows till the middle of the rows and then decrease by 1 till end of the rows.

For example- If start = 4 and repeat = 4 then
  1. start = 4 -> Starting number of pattern,
  2. (repeat * 2) = (4*2)= 8 -> number of rows to print in the pattern 

   4
   55
   666
   7777
   7777
   666
   55
   4

Possible Solution 


 void printPattern(int start, int repeat) {
  int k = repeat;
  for (int i = 0; i < 2 * repeat; i++) { // loop 1 start
   for (int j = 0; (j <= i && j < k); j++) { // loop 2
    System.out.print(start);
   } // end loop 2
   if (i < repeat - 1) {
    start = start + 1;
   }
   if (i > repeat - 1) {
    start = start - 1;
    k = k - 1;
   }
   System.out.println();
  } // end loop 1
 }
Output: 
 
Test 1: (Start = 5 , repeat = 6)

5
66
777
8888
99999
101010101010
101010101010
99999
8888
777
66
5


Test 2: (Start = 3 , repeat = 4)

3
44
555
6666
6666
555
44
3

4 comments :

  1. UnknownOctober 7, 2018 at 12:37 AM

    this question was askend in mindtree 2018
    this pattern printing can also be done in java the code is..

    import java.util.Scanner;
    class Add{
    public static void main(String[] args){
    Scanner ob =new Scanner(System.in);
    System.out.println("enter the starting point ");
    int n=ob.nextInt();
    System.out.print("enter the number of rows ");
    int r=ob.nextInt();
    r=r/2;
    for(int i=1;i<=r;i++){
    for(int j=1;j<=i;j++){
    System.out.print(n);

    }
    n++;
    System.out.println("");
    }n=n-1;



    for(int i=1;i<=r;i++){
    for(int j=i;j<=r;j++){
    System.out.print(n);
    }
    n--;
    System.out.println("");
    }
    }
    }

    ReplyDelete
  2. PriyankaNovember 7, 2018 at 11:28 PM

    #include
    int main()
    {
    int i,j,n,s,p;
    scanf("%d%d",&s,&n);
    for(i=0;i<n;i++){
    for(j=1;j<=i+1;j++){
    p=s+i;
    printf("%d",p);
    }
    printf("\n");
    }
    for(i=0;i<n;i++){
    for(j=1;j<=n-i;j++){
    printf("%d",p-i);
    }
    printf("\n");
    }
    return 0;
    }

    ReplyDelete
  3. UnknownJuly 15, 2019 at 10:27 AM

    pattern 1010101010
    8888
    666
    44
    2 can any one send the program

    ReplyDelete
  4. UnknownAugust 16, 2019 at 8:19 AM

    5
    66
    777
    8888
    8888
    777
    66
    5

    #include

    int main() {
    //code
    int a[10],b[10],i,j,k=5,c;
    for(i=1;i<=4;i++)
    {

    for(j=1;j<=4;j++)
    {
    if(j<=i)
    {
    printf("%d",k);

    }
    else
    printf(" ");

    }
    printf("\n");k++;
    }
    c=k-1;
    for(i=1;i<=4;i++)
    {

    for(j=1;j<=4;j++)
    {
    if(i>=1&&j<=5-i)
    {
    printf("%d",c);

    }
    else
    printf(" ");

    }
    printf("\n");c--;
    }
    }

    ReplyDelete

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