Updation in array

Updation in array

Updation in array is the update operation refers to updating an existing element from the at a given index.This operation is quite similar to the insert method, except that it will replace the existing value at the given index. This means will simply assign a new value at the given index. This method expects two arguments index and value.

 

Diagramatic Representation

 

Algorithm

Consider LA is a linear array with N elements and K is a positive integer such that K<=N.
Following is the algorithm to update an element available at the Kth position of LA.
1. Start
2. Set LA[K-1] = ITEM
3. Stop

 

Updation in array in different languages

Program in C

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void main()
{
int LA[] = {1,3,5,7,8};
int k = 3, n = 5, item = 11;
int i, j;
printf("The original array elements are:\n");
for(i = 0; i<n; i++)
{
printf("LA[%d] = %d \n", i, LA[i]);
}
LA[k-1] = item;
printf("The array elements after updation:\n");
for(i = 0; i<n; i++)
{
printf("LA[%d] = %d \n", i, LA[i]);
}
}
void main() { int LA[] = {1,3,5,7,8}; int k = 3, n = 5, item = 11; int i, j; printf("The original array elements are:\n"); for(i = 0; i<n; i++) { printf("LA[%d] = %d \n", i, LA[i]); } LA[k-1] = item; printf("The array elements after updation:\n"); for(i = 0; i<n; i++) { printf("LA[%d] = %d \n", i, LA[i]); } }
void main()
{
   int LA[] = {1,3,5,7,8};
   int k = 3, n = 5, item = 11;
   int i, j;

   printf("The original array elements are:\n");

   for(i = 0; i<n; i++)
   {
      printf("LA[%d] = %d \n", i, LA[i]);
   }

   LA[k-1] = item;

   printf("The array elements after updation:\n");

   for(i = 0; i<n; i++)
   {
      printf("LA[%d] = %d \n", i, LA[i]);
   }
}

Output

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The original array elements are:
LA[0]=2
LA[1]=4
LA[2]=6
LA[3]=8
LA[4]=9
The array elements after updation:
LA[0]=2
LA[1]=4
LA[2]=11
LA[3]=8
LA[4]=9
The original array elements are: LA[0]=2 LA[1]=4 LA[2]=6 LA[3]=8 LA[4]=9 The array elements after updation: LA[0]=2 LA[1]=4 LA[2]=11 LA[3]=8 LA[4]=9
The original array elements are:
LA[0]=2
LA[1]=4
LA[2]=6
LA[3]=8
LA[4]=9
The array elements after updation:
LA[0]=2
LA[1]=4
LA[2]=11
LA[3]=8
LA[4]=9

Program in CPP

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#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main(){
int arr[3] = {0,1,2};
cout << "Before update "<<arr[2]<<endl;
arr[2]=1;//updating element
cout <<"After update "<<arr[2]<<endl;
}
#include<iostream> #include<bits/stdc++.h> using namespace std; int main(){ int arr[3] = {0,1,2}; cout << "Before update "<<arr[2]<<endl; arr[2]=1;//updating element cout <<"After update "<<arr[2]<<endl; }
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
int main(){
    int arr[3] = {0,1,2};
    cout << "Before update "<<arr[2]<<endl;
    arr[2]=1;//updating element
    cout <<"After update "<<arr[2]<<endl;
}

Output

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Before update 2
After update 1
Before update 2 After update 1
Before update 2
After update 1

 

Program in Java

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class array{
public static void main(String[] args){
int arr[] = {1,2,3,4};
System.out.println("Before update" + arr[2]);
arr[2] = 9;//updating the value
System.out.println("After update" + arr[2]);
}
}
class array{ public static void main(String[] args){ int arr[] = {1,2,3,4}; System.out.println("Before update" + arr[2]); arr[2] = 9;//updating the value System.out.println("After update" + arr[2]); } }
class array{
  public static void main(String[] args){
    int arr[] = {1,2,3,4};
    System.out.println("Before update" + arr[2]);
    arr[2] = 9;//updating the value
    System.out.println("After update" + arr[2]);
  }
}

 

Output

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Before update3
After update9
Before update3 After update9
Before update3
After update9

 

Program in Python

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from collections import deque
def Queries(arr, N, Q):
dq = deque()
for i in range(N):
dq.append(arr[i])
sz = len(Q)
for i in range(sz):
if (Q[i][0] == 0):
front = dq[0]
dq.popleft()
dq.appendleft(front)
elif (Q[i][0] == 1):
back = dq[N - 1]
dq.popleft()
dq.appendleft(back)
elif (Q[i][0] == 2):
dq[Q[i][1]] = Q[i][2]
else:
print(dq[Q[i][1]], end = " ")
if __name__ == '__main__':
arr = [ 1, 2, 3, 4, 5 ]
N = len(arr)
Q = [ [ 0 ], [ 1 ], [ 3, 1 ],
[ 2, 2, 54 ], [ 3, 2 ] ]
Queries(arr, N, Q)
from collections import deque def Queries(arr, N, Q): dq = deque() for i in range(N): dq.append(arr[i]) sz = len(Q) for i in range(sz): if (Q[i][0] == 0): front = dq[0] dq.popleft() dq.appendleft(front) elif (Q[i][0] == 1): back = dq[N - 1] dq.popleft() dq.appendleft(back) elif (Q[i][0] == 2): dq[Q[i][1]] = Q[i][2] else: print(dq[Q[i][1]], end = " ") if __name__ == '__main__': arr = [ 1, 2, 3, 4, 5 ] N = len(arr) Q = [ [ 0 ], [ 1 ], [ 3, 1 ], [ 2, 2, 54 ], [ 3, 2 ] ] Queries(arr, N, Q)
from collections import deque
 
def Queries(arr, N, Q):
 
    dq = deque()
 
    for i in range(N):
        dq.append(arr[i])
 
    sz = len(Q)
 
    for i in range(sz):
 
        if (Q[i][0] == 0):
 
            front = dq[0]
 

            dq.popleft()
 

            dq.appendleft(front)
 

        elif (Q[i][0] == 1):
 
            back = dq[N - 1]
 
            dq.popleft()
 
            dq.appendleft(back)
 
        elif (Q[i][0] == 2):
            dq[Q[i][1]] = Q[i][2]
 
        else:
            print(dq[Q[i][1]], end = " ")
 
if __name__ == '__main__':
 
    arr = [ 1, 2, 3, 4, 5 ]
    N = len(arr)
 
    Q = [ [ 0 ], [ 1 ], [ 3, 1 ],
          [ 2, 2, 54 ], [ 3, 2 ] ]
 
    Queries(arr, N, Q)

Output

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2 54
2 54
2 54

 

Program in Javascript

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<script>
function Queries(arr,N,Q)
{
let dq = [];
for (let i = 0; i < N; i++)
{
dq.push(arr[i]);
}
let sz = Q.length;
for (let i = 0; i < sz; i++)
{
if (Q[i][0] == 0)
{
let front = dq[0];
dq.shift();
dq.push(front);
}
else if (Q[i][0] == 1)
{
let back = dq[dq.length - 1];
dq.pop();
dq.unshift( back);
}
else if (Q[i][0] == 2)
{
dq[Q[i][1]] = Q[i][2];
}
else
{
document.write(dq[Q[i][1]] + " ");
}
}
}
let arr=[1, 2, 3, 4, 5];
let N = arr.length;
let Q = [[0], [1], [3, 1],
[2, 2, 54], [3, 2]];
Queries(arr, N, Q);
</script>
<script> function Queries(arr,N,Q) { let dq = []; for (let i = 0; i < N; i++) { dq.push(arr[i]); } let sz = Q.length; for (let i = 0; i < sz; i++) { if (Q[i][0] == 0) { let front = dq[0]; dq.shift(); dq.push(front); } else if (Q[i][0] == 1) { let back = dq[dq.length - 1]; dq.pop(); dq.unshift( back); } else if (Q[i][0] == 2) { dq[Q[i][1]] = Q[i][2]; } else { document.write(dq[Q[i][1]] + " "); } } } let arr=[1, 2, 3, 4, 5]; let N = arr.length; let Q = [[0], [1], [3, 1], [2, 2, 54], [3, 2]]; Queries(arr, N, Q); </script>
<script>

function Queries(arr,N,Q)
{

  let dq = [];
  
  for (let i = 0; i < N; i++)
  {
    dq.push(arr[i]);
  }
  
  let sz = Q.length;
  
  for (let i = 0; i < sz; i++)
  {
    if (Q[i][0] == 0)
    {

      let front = dq[0];
  
      dq.shift();
  
      dq.push(front);
    }
  
    else if (Q[i][0] == 1)
    {

      let back = dq[dq.length - 1];
  
      dq.pop();
  
      dq.unshift( back);
    }
  

    else if (Q[i][0] == 2)
    {
      dq[Q[i][1]] = Q[i][2];
    }
  
    
    else
    {
      document.write(dq[Q[i][1]] + " ");
    }
  }
}
 
let arr=[1, 2, 3, 4, 5];
let N = arr.length;
    
 
let Q = [[0], [1], [3, 1],
[2, 2, 54], [3, 2]];
Queries(arr, N, Q);
 
 
</script>

Output

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2 54
2 54
2 54

 

Time Complexity:

O(n)

 

Array operation

Insertion in array

Deletion in array

Traversing in array

Searching in array

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