print repeated array of 2D

[helnow]

I need help with Printing duplicate 2D array elements

if for 1D array

for (int i = 0; i<0; i++){
     for (int j = i+1; j<0; i++)
{
      if (arra[i] == arra[j]){
          cout<<arra[i]<<endl
}
}
}

how would i do for 2D

any help is appreciated

Thank you

[JLBorges]

#include <iostream>

int main()
{
    const int M = 3, N = 5 ;

    int a[M][N] = { { 0, 1, 2, 3, 4, }, { 5, 2, 6, 4, 7 }, { 6, 7, 5, 8, 9 } };
    
    for( int i = 0 ; i < M*N ; ++i )
    {
        for( int j = i+1 ; j < M*N ; ++j )
            if( a[0][i] == a[0][j] ) std::cout << a[0][i] << ' ' ; 
    }
    std::cout << '\n' ;
}

http://coliru.stacked-crooked.com/a/945d1fc2dabfa5fa

[helnow]

Thanks, i see what you did, but what if i was doing 10x10 array,
and what if i want to print out the indexes that have the same number as well

for example
if index [1][0] and index[5][8] have the same number,
and i want to print out the position of the indexes and as well as the number
they have in common with

[JLBorges]

Since we know the extents (number of rows and columns) of the array, the indices can be computed.

#include <iostream>

int main()
{
    const int NROWS = 3, NCOLS = 5 ;

    int a[NROWS][NCOLS] = { { 0, 1, 2, 3, 4, }, { 5, 2, 6, 4, 7 }, { 6, 7, 5, 8, 4 } };

    for( int i = 0 ; i < NROWS*NCOLS ; ++i )
    {
        const int row = i / NCOLS ;
        const int col = i % NCOLS ;

        for( int j = i+1 ; j < NROWS*NCOLS ; ++j )
        {
            const int row2 = j / NCOLS ;
            const int col2 = j % NCOLS ;
            
            if( a[row][col] == a[row2][col2] )
                std::cout << "a[" << row << "][" << col << "]  " << a[row][col]
                           << "  a[" << row2 << "][" << col2 << "]\n" ;
        }
    }
}

http://coliru.stacked-crooked.com/a/0269d82e42280538

[helnow]

Thanks a lot
it is also to do it in a four nested loops right?
because i did that at first, but wasn't sure if it was correct, but i got similar answer with your code and four nested loops

Thanks again for your help

[JLBorges]

> it is also to do it in a four nested loops right?

Yes.