What I know is Linear Search and Sequential Search are same, if my knowledge is correct then kindly guide me wether the below algorithm work for linear search or not as it is a Sequential Search algorithm.

Step 1. Initialize searcharray, searchno, length.
Step 2. Initialize pos=0.
Step 3. Repeat step 4 till pos<=length.
Step 4. if searcharray[pos]=searchno
return pos
else
increment pos by 1.

Secondly I need help on how can I move the found value at the first element of an array, I have this code which can find value which user enter in an array using linear search.

#include <iostream.h>
#include<conio.h>
int linearSearch(const int a[], int size, int key);
int main() {
   const int SIZE = 8;
   int a1[SIZE] = {8, 4, 5, 3, 2, 9, 4, 1};

   cout << linearSearch(a1, SIZE, 8) << endl;  // 0
   cout << linearSearch(a1, SIZE, 4) << endl;  // 1
   cout << linearSearch(a1, SIZE, 99) << endl; // 8 (not found)
}

// Search the array for the given key
// If found, return array index [0, size-1]; otherwise, return size
int linearSearch(const int a[], int size, int key) {
   for (int i = 0; i < size; ++i) {
      if (a[i] == key) return i;
   }
   return size;
}

Step 2. Initialize pos=0.
Step 3. Repeat step 4 till pos<=length.

When starting at 0, that implies a half open range. Step 3 is wrong in that it says to repeat until pos is less than or equal to length. In other words, the loop will never run because pos will never be less than length.

Instead, step 3 should read "Repeat step 4 until pos = length". This stops the loop when pos is equal to length so that the array isn't indexed out of bounds.

Secondly I need help on how can I move the found value at the first element of an array

So you want to move the located value to the first element? What happens to the other elements? Typically there are two approaches to a move-first algorithm:

  1. Swap the current first with the found value.

    for (i = 0; i < n; i++) {
        if (is_match(a[i])) {
            break;
        }
    }
    
    if (i != 0) {
        swap(a[0], a[i]);
    }
    
  2. Shift all values to fill in the vacated hole, then copy the value to the first position.

    for (i = 0; i < n; i++) {
        if (is_match(a[i])) {
            break;
        }
    }
    
    if (i != 0) {
        save = a[i];
    
        while (i > 0) {
            a[i] = a[i - 1];
            --i;
        }
    
        a[0] = save;
    }
    
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