# Replace every element with the smallest of the others

• Difficulty Level : Easy
• Last Updated : 16 Jun, 2021

Given an array arr[] of N distinct integers, the task is to replace every element by smallest among all other remaining elements of the array.
Examples:

Input: arr[] = { 4, 2, 1, 3 }
Output: arr[]= { 1, 1, 2, 1 }
Explanation:
For arr[0], the smallest out of the remaining elements is 1. So arr[0] is replaced by 1.
For arr[1], the smallest out of the remaining elements is 1. So arr[1] is replaced by 1.
For arr[2], the smallest out of the remaining elements is 2. So arr[2] is replaced by 2.
For arr[3], the smallest out of the remaining elements is 1. So arr[3] is replaced by 1.
Input: arr[] = { 1, 5, 2, 4 }
Output: arr[] = { 2, 1, 1, 1 }

Naive Approach:
Run a nested loop and find the smallest of all other elements for every single element and store them.
Time Complexity: O(N2)
Efficient Approach:

1. Traverse the array and find the two smallest elements of the array.
2. Traverse the array again, now replace the smallest element with the second smallest element and replace every other element in the array with the smallest element.
3. Print the modified array.

Below is the implementation of above approach:

## C++

 `// C# implementation to find the``// maximum array sum by concatenating``// corresponding elements of given two arrays``using` `System;``class` `GFG{` `// Function to join the two numbers``static` `int` `joinNumbers(``int` `numA, ``int` `numB)``{``    ``int` `revB = 0;` `    ``// Loop to reverse the digits``    ``// of the one number``    ``while` `(numB > 0)``    ``{``        ``revB = revB * 10 + (numB % 10);``        ``numB = numB / 10;``    ``}` `    ``// Loop to join two numbers``    ``while` `(revB > 0)``    ``{``        ``numA = numA * 10 + (revB % 10);``        ``revB = revB / 10;``    ``}` `    ``return` `numA;``}` `// Function to find the maximum array sum``static` `int` `findMaxSum(``int` `[]A, ``int` `[]B, ``int` `n)``{``    ``int` `[]maxArr = ``new` `int``[n];` `    ``// Loop to iterate over the two``    ``// elements of the array``    ``for``(``int` `i = 0; i < n; ++i)``    ``{``        ``int` `X = joinNumbers(A[i], B[i]);``        ``int` `Y = joinNumbers(B[i], A[i]);``        ``int` `mx = Math.Max(X, Y);``    ` `        ``maxArr[i] = mx;``    ``}` `    ``// Find the array sum``    ``int` `maxAns = 0;``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ``maxAns += maxArr[i];``    ``}` `    ``// Return the array sum``    ``return` `maxAns;``}` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``int` `N = 5;``    ``int` `[]A = { 11, 23, 38, 43, 59 };``    ``int` `[]B = { 36, 24, 17, 40, 56 };` `    ``Console.WriteLine(findMaxSum(A, B, N));``}``}` `// This code is contributed by Rajput-Ji`

## Java

 `// Java code for the above approach.``import` `java.util.*;` `class` `GFG {` `    ``static` `void` `ReplaceElements(``        ``int``[] arr, ``int` `n)``    ``{` `        ``// There should be atleast``        ``// two elements``        ``if` `(n < ``2``) {` `            ``System.out.println(``                ``"Invalid Input"``);` `            ``return``;``        ``}` `        ``int` `firstSmallest``            ``= Integer.MAX_VALUE;``        ``int` `secondSmallest``            ``= Integer.MAX_VALUE;` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// If current element is smaller``            ``// than firstSmallest then update``            ``// both firstSmallest``            ``// and secondSmallest``            ``if` `(arr[i] < firstSmallest) {` `                ``secondSmallest = firstSmallest;``                ``firstSmallest = arr[i];``            ``}` `            ``// If arr[i] is in between``            ``// firstSmallest and secondSmallest``            ``// then update secondSmallest``            ``else` `if` `(arr[i] < secondSmallest``                     ``&& arr[i] != firstSmallest)``                ``secondSmallest = arr[i];``        ``}` `        ``// Replace every element by``        ``// smallest of all other elements``        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(arr[i] != firstSmallest)``                ``arr[i] = firstSmallest;` `            ``else``                ``arr[i] = secondSmallest;``        ``}` `        ``// Print the modified array.``        ``for` `(``int` `i = ``0``; i < n; ++i) {``            ``System.out.print(arr[i]``                             ``+ ``", "``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(``        ``String[] args)``    ``{``        ``int` `arr[] = { ``4``, ``2``, ``1``, ``3` `};``        ``int` `n = arr.length;` `        ``ReplaceElements(arr, n);``    ``}``}`

## Python3

 `# Python3 code for the above approach.``def` `ReplaceElements(arr, n):` `    ``# There should be atleast``    ``# two elements``    ``if` `(n < ``2``):``        ``print``(``"Invalid Input"``)``        ``return` `    ``firstSmallest ``=` `10` `*``*` `18``    ``secondSmallest ``=` `10` `*``*` `18` `    ``for` `i ``in` `range``(n):` `        ``# If current element is smaller``        ``# than firstSmallest then update``        ``# both firstSmallest``        ``# and secondSmallest``        ``if` `(arr[i] < firstSmallest):``            ``secondSmallest ``=` `firstSmallest``            ``firstSmallest ``=` `arr[i]` `        ``# If arr[i] is in between``        ``# firstSmallest and secondSmallest``        ``# then update secondSmallest``        ``elif` `(arr[i] < secondSmallest ``and``              ``arr[i] !``=` `firstSmallest):``            ``secondSmallest ``=` `arr[i]` `    ``# Replace every element by``    ``# smallest of all other elements``    ``for` `i ``in` `range``(n):` `        ``if` `(arr[i] !``=` `firstSmallest):``            ``arr[i] ``=` `firstSmallest` `        ``else``:``            ``arr[i] ``=` `secondSmallest` `    ``# Print the modified array.``    ``for` `i ``in` `arr:``        ``print``(i, end ``=` `", "``)`  `# Driver code``if` `__name__ ``=``=` `'__main__'``:``     ` `    ``arr``=` `[ ``4``, ``2``, ``1``, ``3` `]``    ``n ``=` `len``(arr)` `    ``ReplaceElements(arr, n)` `# This code is contributed by mohit kumar 29`

## C#

 `// C# code for the above approach.``using` `System;` `class` `GFG{` `static` `void` `ReplaceElements(``int``[] arr, ``int` `n)``{` `    ``// There should be atleast``    ``// two elements``    ``if` `(n < 2)``    ``{``        ``Console.Write(``"Invalid Input"``);` `        ``return``;``    ``}` `    ``int` `firstSmallest = Int32.MaxValue;``    ``int` `secondSmallest = Int32.MaxValue;` `    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ` `       ``// If current element is smaller``       ``// than firstSmallest then update``       ``// both firstSmallest``       ``// and secondSmallest``       ``if` `(arr[i] < firstSmallest)``       ``{``           ``secondSmallest = firstSmallest;``           ``firstSmallest = arr[i];``       ``}``       ` `       ``// If arr[i] is in between``       ``// firstSmallest and secondSmallest``       ``// then update secondSmallest``       ``else` `if` `(arr[i] < secondSmallest &&``                ``arr[i] != firstSmallest)``           ``secondSmallest = arr[i];``    ``}` `    ``// Replace every element by``    ``// smallest of all other elements``    ``for``(``int` `i = 0; i < n; i++)``    ``{``       ``if` `(arr[i] != firstSmallest)``           ``arr[i] = firstSmallest;``       ``else``           ``arr[i] = secondSmallest;``    ``}` `    ``// Print the modified array.``    ``for``(``int` `i = 0; i < n; ++i)``    ``{``       ``Console.Write(arr[i] + ``", "``);``    ``}``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `[]arr = { 4, 2, 1, 3 };``    ``int` `n = arr.Length;` `    ``ReplaceElements(arr, n);``}``}` `// This code is contributed by Nidhi_Biet`

## Javascript

 ``

Output:

`1, 1, 2, 1,`

Time Complexity: O(N)
Space Complexity: O(1)

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