NPTEL - WEEK 5 - ASSIGNMENT 5 [ QUESTION 02 ] SOLUTION

WEEK 5 ASSIGNMENT 5 QUESTION 02 SOLUTION 

QUESTION : 

Consider a matrix M of integers. Divide M into 4 sub-matrices. These sub-matrices are called as Quadrants. Report the Quadrant number which has the smallest minimum-element. If two or more quadrants have same smallest minimum, report the smallest quadrant index.
The matrix M is divided into four quadrants by halving the rows and columns. If row/column is an odd number, divide them in such a way that the first half of the row/column should be one smaller than the second half.

The four quadrants are numbered from 1 to 4 in the structure shown below: Q1 | Q2
---+---
 Q3 | Q4
INPUT FORMAT: M is a matrix of integers. You would be given two numbers m and n specifying the number of rows and columns. This would be followed by m lines of n integers each specifying the data of the m*n matrix M.
n and m will be greater than 1 and less than 12. OUTPUT FORMAT: Print in a separate line, the quadrant number with the smallest minimum-element.

SOLUTION : 

#include<stdio.h>
int smallel(int *ptr, int len);
int main()
{
    int m, n, p, q, x, y, i,k, j, index;
    int  M[100][100];
    int  quad1[10], quad2[10], quad3[10], quad4[10];
    int  s[4], small;

    scanf("%d %d", &m, &n);

    M[m][n];

    for(i=0; i<m; i++)
    {
        for(j=0; j<n; j++)
        {
            scanf("%d", &M[i][j]);
        }
    }

    // Finding the quadrants -->
    // 1 quadrant.
    k=0;
    p=0;
    for(i=0; i<m/2; i++)
    {
        for(j=0; j<n/2; j++)
        {
            quad1[k] = M[i][j];
            k++;
            p++;

        }
    }

    // 2nd quadrant.
    k=0;
    q=0;
    for(i=0; i<m/2; i++)
    {
        for(j=n/2; j<n; j++)
        {
            quad2[k] = M[i][j];
            k++;
            q++;
        }
    }

    // 3rd quadrant.
    k=0;
    x = 0;
    for(i=m/2; i<m; i++)
    {
        for(j=0; j<n/2; j++)
        {
            quad3[k] = M[i][j];
            k++;
            x++;
        }
    }

    // 4 quadrant.
    k=0;
    y = 0;
    for(i=m/2; i<m; i++)
    {
        for(j=n/2; j<n; j++)
        {
            quad4[k] = M[i][j];
            k++;
            y++;
        }
    }

    // Find the small element in every quadrant.
    s[0] = smallel(quad1, p);
    s[1] = smallel(quad2, q);
    s[2] = smallel(quad3, x);
    s[3] = smallel(quad4, y);

    // Find the small element in s array.
    small = s[0];
    index = 1;
    for(i=0; i<4; i++)
    {
        if(small > s[i])
        {
            small = s[i];
            index = i+1;
        }
    }
   
    printf("%d", index);
   
    return 0;
}

int smallel(int *ptr, int len)
{
    int small = *ptr;
    for(int i=0; i<len; i++)
    {
        if(small > *(ptr+i))
        {
            small = *(ptr+i);
        }
    }
    return small;
}

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