Print pair with maximum AND value in an array in C Program.

CServer Side ProgrammingProgramming

According to the problem we are given an array of n positive integers, we have to find a pair with maximum AND value from the array.

Example

Input: arr[] = { 4, 8, 12, 16 }
Output: pair = 8 12
The maximum and value= 8

Input:arr[] = { 4, 8, 16, 2 }
Output: pair = No possible AND
The maximum and value = 0

For finding Maximum AND value is similar to find out Maximum AND value in an array. Program must find out the pair of elements resulting in obtained AND value. For finding the elements, simply traverse the whole array and find the AND value of each element with the obtained maximum AND value (result) and if arr[i] & result == result, that means arr[i] is the element which will generate maximum AND value. Also, in the case if maximum AND value (result) is zero then we should print “Not possible” in that case.

Algorithm

int checkBit(int pattern, int arr[], int n)
START
STEP 1: DECLARE AND INITIALIZE count AS 0
STEP 2: LOOP FOR i = 0 AND i < n AND i++
   IF (pattern & arr[i]) == pattern THEN,
      INCREMENT count BY 1
STEP 3: RETURN count
STOP
int maxAND(int arr[], int n)
START
STEP 1: DECLARE AND INITIALIZE res = 0 AND count
STEP 2: LOOP FOR bit = 31 AND bit >= 0 AND bit--
   count = GOTO FUNCTION checkBit(res | (1 << bit), arr,n)
   IF count >= 2 THEN,
      res |= (1 << bit);
   END IF
   IF res == 0
      PRINT "no possible AND”
   ELSE
      PRINT "Pair with maximum AND= "
   count = 0;
   LOOP FOR int i = 0 AND i < n && count < 2 AND i++
      IF (arr[i] & res) == res THEN,
         INCREMENT count BY 1
         PRINT arr[i]
      END IF
   END FOR
END FOR
RETURN res
STOP

Example

#include <stdio.h>
int checkBit(int pattern, int arr[], int n){
   int count = 0;
   for (int i = 0; i < n; i++)
      if ((pattern & arr[i]) == pattern)
         count++;
   return count;
}
// Function for finding maximum AND value pair
int maxAND(int arr[], int n){
   int res = 0, count;
   for (int bit = 31; bit >= 0; bit--) {
      count = checkBit(res | (1 << bit), arr, n);  
      if (count >= 2)
         res |= (1 << bit);
   }
   if (res == 0) //if there is no pair available
      printf("no possible and\n");
   else { //Printing the pair available
      printf("Pair with maximum AND= ");
      count = 0;
      for (int i = 0; i < n && count < 2; i++) {
         // incremnent count value after
         // printing element
         if ((arr[i] & res) == res) {
            count++;
            printf("%d ", arr[i]);
         }
      }
   }
   return res;
}
int main(int argc, char const *argv[]){
   int arr[] = {5, 6, 2, 8, 9, 12};
   int n = sizeof(arr)/sizeof(arr[0]);
   int ma = maxAND(arr, n);
   printf("\nThe maximum AND value= %d ", ma);
   return 0;
}

Output

If we run the above program then it will generate the following output −

pair = 8 9
The maximum and value= 8
raja
Published on 22-Aug-2019 08:45:50
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