Program to find repeated number in array

Logic to find repeated number in array

compare array element

array[0] == array[1]
array[0] == array[2]
array[0] == array[3]
array[0] == array[4]


array[1] == array[0]
array[1] == array[2]
array[1] == array[3]
array[1] == array[4]


array[2] == array[0]
array[2] == array[1]
array[2] == array[3]
array[2] == array[4]


array[3] == array[0]
array[3] == array[1]
array[3] == array[2]
array[3] == array[4]


array[4] == array[0]
array[4] == array[1]
array[4] == array[2]
array[4] == array[3]

Code:

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#include <stdio.h>


int main ()
{
    int i, array[5] = {1, 2, 3, 4, 4};
   
    int duplicate = array[0];


    for ( i = 1; i <5; i++)
       
        {
          if (duplicate ==array[i]
          {
              printf("%d ", duplicate);
          }
        }
           
    return 0;
}

How to check all conditions

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Perhaps make your code follow your logic, which suggests you need two nested loops.

Quote:

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Originally Posted by Salem

Perhaps make your code follow your logic, which suggests you need two nested loops.

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I'm trying to implement logic. I understand nested loops.

Code:

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#include <stdio.h>

int main(void)
 {
       
        int i, j;
       
        for( i = 0; i < 5; i++)
        {
          printf(" i = %d \n", i);
       
                  for( j = 0; j < 5; j++)
                      {
                            printf(" j = %d \n",j); 
                        }         
         
        }
       
        return 0;
}

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I'm sure you'll figure it out if you also print array[i] and array[j]

Quote:

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Originally Posted by Salem

I'm sure you'll figure it out if you also print array[i] and array[j]

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What's wrong here

Code:

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#include <stdio.h>
int main(void)
 {
       
        int i, j;
        int array[5] = {1, 2, 3, 4, 4};
       
        for( i = 0; i < 5; i++)
        {
       
                  for( j = 0; j < 5; j++)
                      {
                            if (array[j] ==array[j+1])
                            {
                                printf("%d", array[j] );
                                break;
                            }
                        }         
           
        }
       
        return 0;
}

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> What's wrong here
Did you do as I suggested, or did you lunge straight for the problem without understanding intermediate steps?

Quote:

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Originally Posted by Salem

> What's wrong here
Did you do as I suggested, or did you lunge straight for the problem without understanding intermediate steps?

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When I run that program it find repeated number but it print that number five times like 44444

I just want to find repeat number and that is 4

So what happened to the "int duplicate = array[0];" idea, except you now make it int duplicate = array[i];

Quote:

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Originally Posted by Salem

So what happened to the "int duplicate = array[0];" idea, except you now make it int duplicate = array[i];

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This program find repeated number

Code:

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#include <stdio.h>

int main(void)
 {
       
        int i, j;
               
        int array[5] = {1, 1, 3, 4, 4};
       
        for( i = 0; i < 5; i++)
        {
       
                  for( j = i+1; j < 5; j++)
                                         
                                            if (array[i] == array[j])
                            {
                                                                printf("%d ", array[i] );                                                                                                                               
                                                        }                                                         
        }
     
        return 0;
}

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I guess you should break out of the loop as soon as it finds the repeated number

Code:

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if(array[i] == array[j])
{
    printf("Found repeated number %d\n\n", array[i]);
    break;
}

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