Find maximum value in an array by recursion

Question

// Find a maximum element in the array.
findMax(A)
   findMaxHelper(A, 0, A.length)

findMaxHelper(A, left, right)
   if (left == right - 1) 
      return A[left]
   else
      max1 = findMaxHelper(A, left, (right + left) / 2)
      max2 = findMaxHelper(A, (right + left) / 2, right)

      if (max1 > max2) 
         return max1 
      else 
         return max2

I am having a hard time understanding what is happening in this pseudo-code.

Can someone help explain what is happening at each line. I need to understand this code before I can answer the questions.

I know that the function findMax calls the helper function findMaxHelper, then findMaxHelper uses recursion. Other than that, I really don't understand it.

Well, one thing that is happening is that the max element of the array is being calculated in a very expensive way! — Gene, May 28 '13 at 23:57

Jainendra · Answer 1 · 2019-09-17T13:58:08.557

30

You are using Divide and Conquer algorithm for finding the maximum element from the array. First, you are dividing the array into individual elements (divide), then you are comparing the elements (conquer). You are dividing the array using calling findMaxHelper recursively.

The general idea of Divide and conquer is shown in the figure:

enter image description here

Example:

enter image description here Here max is same as your findMaxHelper function with two arguments i.e. left and right.

Check this example for more in depth understanding of the concept.

edited Sep 17 '19 at 13:58

answered Oct 29 '12 at 07:09

Jainendra

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@JustinBains `left` and `right` are the indexes of the first and last element of the arrays(Initial as well as intermediate arrays). – Jainendra Oct 29 '12 at 07:17
2

A general suggestion to anyone struggling with understanding recursive code: do not try to dive deep and follow. Better do a "zoom out" and try to understand the bigger picture. Recursive functions usually take the input, perform basic operation and **repeat the same for a smaller problem**, just like in this code snippet. You should try to identify the smaller problem(s), that's the core of understanding such code. – SomeWittyUsername Oct 29 '12 at 07:19

score 2 · Answer 2 · answered Jun 01 '13 at 16:18

Jaguar has put the concept quite nicely and Paul has provided correct and detailed explanation. To add to this , I would like to share a simple C code that gives you an idea how the code gets executed. Here's the code with the same input Jaguar used :

#include<stdio.h>
int findMaxHelper(int A[], int left, int right){
   int max1,max2;
   int static tabcount;
   int loop;
   for(loop = 0 ; loop <tabcount;loop++) printf("\t");
   tabcount++;
   printf(" Entering: findMaxHelper(A, left = %d ,right = %d)\n\n",left,right);
   if (left == right - 1){ 
      for(loop = 0 ; loop <tabcount;loop++) printf("\t");
      printf("\b\b\b\b\b\b\bLeaving: findMaxHelper(A, left = %d ,right = %d)| returning %d\n\n",left,right , A[left]);
      tabcount--;
      return A[left];
   }
   else
   {
      max1 = findMaxHelper(A, left, (right + left) / 2);
      max2 = findMaxHelper(A, (right + left) / 2, right);

      if (max1 > max2){ 
    for(loop = 0 ; loop <tabcount;loop++) printf("\t");
    printf("\b\b\b\b\b\b\bLeaving: findMaxHelper(A, left = %d ,right = %d) | returning max1=%d\n\n",left,right,max1);
    tabcount--;
    return max1;
    }
      else {
     for(loop = 0 ; loop <tabcount;loop++) printf("\t");
     printf("\b\b\b\b\b\b\bLeaving: findMaxHelper(A, left = %d ,right = %d)| returning max2=%d\n\n",left,right,max2);
     tabcount--;
     return max2;
    }

   }
}

int main (){
    int A[] = { 34,3,47,91,32,0 };
    int Ans =findMaxHelper(A,0,7);  
    printf( "And The Answer Is = %d \n",Ans);
}

U can copy paste the code on ur linux machine ...Maybe put sleep(5) after every printf and see how recursion ACTUALLY works !... Hope this helps... I will also share the output from my system here :

Entering: findMaxHelper(A, left = 0 ,right = 7)

     Entering: findMaxHelper(A, left = 0 ,right = 3)

         Entering: findMaxHelper(A, left = 0 ,right = 1)

         Leaving: findMaxHelper(A, left = 0 ,right = 1)| returning 34

         Entering: findMaxHelper(A, left = 1 ,right = 3)

             Entering: findMaxHelper(A, left = 1 ,right = 2)

             Leaving: findMaxHelper(A, left = 1 ,right = 2)| returning 3

             Entering: findMaxHelper(A, left = 2 ,right = 3)

             Leaving: findMaxHelper(A, left = 2 ,right = 3)| returning 47

         Leaving: findMaxHelper(A, left = 1 ,right = 3)| returning max2=47

     Leaving: findMaxHelper(A, left = 0 ,right = 3)| returning max2=47

     Entering: findMaxHelper(A, left = 3 ,right = 7)

         Entering: findMaxHelper(A, left = 3 ,right = 5)

             Entering: findMaxHelper(A, left = 3 ,right = 4)

             Leaving: findMaxHelper(A, left = 3 ,right = 4)| returning 91

             Entering: findMaxHelper(A, left = 4 ,right = 5)

             Leaving: findMaxHelper(A, left = 4 ,right = 5)| returning 32

         Leaving: findMaxHelper(A, left = 3 ,right = 5) | returning max1=91

         Entering: findMaxHelper(A, left = 5 ,right = 7)

             Entering: findMaxHelper(A, left = 5 ,right = 6)

             Leaving: findMaxHelper(A, left = 5 ,right = 6)| returning 0

             Entering: findMaxHelper(A, left = 6 ,right = 7)

             Leaving: findMaxHelper(A, left = 6 ,right = 7)| returning 0

         Leaving: findMaxHelper(A, left = 5 ,right = 7)| returning max2=0

     Leaving: findMaxHelper(A, left = 3 ,right = 7) | returning max1=91

 Leaving: findMaxHelper(A, left = 0 ,right = 7)| returning max2=91

And The Answer Is = 91

score 0 · Answer 3 · answered Oct 29 '12 at 07:16

0

findMaxHelper divides the array into half each time, and find the max in left,right:

eg you have array A = [1, 3, 5, 8], call findMax(A) -> findMaxHelper(A, 0, A.length):

     max1 | max2
     1 3  | 5 8

max1|max2 | max1|max2
1   |3    | 5   |8

answered Oct 29 '12 at 07:16

Mifeng

1,525
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score -1 · Answer 4 · edited Nov 17 '16 at 06:14

-1

#include<stdio.h>
#include<stdlib.h>

int high,*a,i=0,n,h;
int max(int *);

int main()
{

    printf("Size of array: ");
    scanf("%d",&n);

    a=(int *)malloc(n*sizeof(int));         //dynamic allocation
    for(i=0;i<n;i++)
    {
        scanf("%d",(a+i));
    }
        i=0;
    high=*a;
    h=max(a);
    printf("The highest element is %d\n",h);
}

int max(int *a)
{

    if(i<n)
    {   
        if(*(a+i)>high)
        {high=*(a+i);}
    i++;
    max(a);                     //recursive call
    }

    return high;
}

edited Nov 17 '16 at 06:14

sunkuet02

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answered Nov 17 '16 at 04:56

Praveen Kumar Sinha

1

1

Welcome to SO. Note that the OP was really asking for an explanation of the psuedo-code. Including an answer that is code with no explanation is unlikely to be useful. – sprinter Nov 17 '16 at 05:25

score -2 · Answer 5 · answered Feb 22 '16 at 00:19

Basically finding max in array is not recommended by recursion as it is not required. Divide and conquer algorithms(recursive) are more time costly. But even though if you want to use it, you can use my below algorithm. Basically, it brings the largest element of array at first position and has almost linear running time.(This algo is just a recursive-illusion though!):

        int getRecursiveMax(int arr[], int size){
          if(size==1){
                      return arr[0];
          }else{
                 if(arr[0]< arr[size-1]){
                                      arr[0]=arr[size-1];
                     }
                 return(getRecursiveMax(arr,size-1));
            }

          }

Find maximum value in an array by recursion

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