# Number pattern program print in down up approach

Number pattern program print in down up approach

## Logic:

- Get the row size from the user.
- The below steps are executed for
**(2*n)-1**times. - Determine the starting position to store the count value.
- Then decrement row value and increment column value.
- At last, print the array to display patterns.

## Explanation:

- a=>stores the pattern value, n=>number of rows, k=>determines the row, l=>determines the column and cnt=>value to store
- Let’s consider input n=4 and cnt=1
- By using n we can find how many times the loop going to execute which is (2*n)-1 times ie, 7 times
- Store the cnt=1 value into the array
**a**into the position a[k][l] now k=0, m=0,l=0, and cnt=1 and m is used only for upper loop count to execute - For the next row k=1, l=0, m=1 and cnt=2. Now inner loop executes for two times and then l increments and k value decrements now k=0 and l=1, cnt=3 value stores into the array
- This process continues for the upper half of the pattern and cnt=11
- Then moves to the lower half where we increment both row and column
- Now l=1, k=3, m=3, cnt=11, and the same thing going on loop executes for 3 times, and count store into the array
- Then l=2, k=3, m=3, cnt=14, and the same thing going on loop executes for 2 times, and count store into the array
- Then l=3, k=3, m=3, cnt=16, and the same thing going on loop executes for 1 time, and count store into the array
- And exits from the outer loop then prints the pattern

## Program:

#include<stdio.h> void main() { /* a->stores the pattern value n->number of rows k->determines the row l->determines the column cnt->value to store */ int a[10][10],n,i,j,k,cnt=1,l,m; printf("Enter the number:"); scanf("%d",&n); for(i=0;i<(2*n)-1;i++) //loops executes for (2*n)-1 times { k=i,m=i; l=0; if(i>=n) //applies only for lower half { l=(i+1)%n; k=n-1; m=n-1; } for(j=l;j<=m;j++) { a[k][j]=cnt++; //store the count into array k--; } } for(i=0;i<n;i++) //prints the pattern { for(j=0;j<n;j++) { printf("%d ",a[i][j]); } printf("\n"); } }