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Rearrange Array to maximize number having Array elements as digits based on given conditions
• Last Updated : 29 Sep, 2020

Given an array of integers arr[] and a binary string str of length N, the task is to rearrange given array by swapping array elements from indices having the same character in the string, such that the number formed by the elements of the rearranged array as digits is the maximum possible.

Examples:

Input: arr[]={1, 3, 4, 2}, str=”0101”
Output: 4 3 1 2
Explanation:
Since arr is less than arr, so swap them. Therefore the maximum possible number from the array is 4, 3, 1, 2.

Input: arr[] = { 1, 3, 456, 6, 7, 8 }, str = “101101”
Output: 8 7 6 456 3 1
Explanation:
Array elements present at 0-chractered indices: {3, 7}
Largest number that can be formed from the above two numbers is 73
Array elements present at 1-chractered indices: {1, 456, 6, 8}
Largest number that can be formed from the above two numbers is 864561
Therefore, maximum number that can be generated from the array is 87645631

Approach: Follow the steps below to solve the problem:

1. Create two arrays to store 0-charactered index elements and 1-charactered index elements from the array.
2. Sort the arrays to form largest possible numbers from these two arrays.
3. Iterate over str and based on the characters, place array elements from the sorted arrays.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``using` `namespace` `std;` `// Comparison Function to sort()``int` `myCompare(``int` `a, ``int` `b)``{``    ``string X = to_string(a);``    ``string Y = to_string(b);` `    ``// Append Y at the end of X``    ``string XY = X.append(Y);` `    ``// Append X at the end of Y``    ``string YX = Y.append(X);` `    ``// Compare and return greater``    ``return` `XY.compare(YX) < 0 ? 1 : 0;``}` `// Function to return the rearranged``// array in the form of largest``// possible number that can be formed``void` `findMaxArray(vector<``int``>& arr, string& str)``{``    ``int` `N = arr.size();``    ``vector<``int``> Z, O, ans(N);` `    ``for` `(``int` `i = 0; i < N; i++) {``        ``if` `(str[i] == ``'0'``) {``            ``Z.push_back(arr[i]);``        ``}` `        ``else` `{``            ``O.push_back(arr[i]);``        ``}``    ``}` `    ``// Sort them in decreasing order``    ``sort(Z.rbegin(), Z.rend(), myCompare);``    ``sort(O.rbegin(), O.rend(), myCompare);` `    ``int` `j = 0, k = 0;` `    ``// Generate the sorted array``    ``for` `(``int` `i = 0; i < N; i++) {``        ``if` `(str[i] == ``'0'``) {``            ``ans[i] = Z[j++];``        ``}``        ``else` `{``            ``ans[i] = O[k++];``        ``}``    ``}` `    ``for` `(``int` `i = 0; i < N; i++) {``        ``cout << ans[i] << ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``vector<``int``> arr = { 1, 3, 456, 6, 7, 8 };``    ``string str = ``"101101"``;``    ``findMaxArray(arr, str);``    ``return` `0;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;``import` `java.lang.*;` `class` `GFG{` `// Function to return the rearranged``// array in the form of largest``// possible number that can be formed``static` `void` `findMaxArray(``int``[] arr, String str)``{``    ``int` `N = arr.length;``    ``ArrayList Z = ``new` `ArrayList<>(),``                       ``O = ``new` `ArrayList<>();``                       ` `    ``int``[] ans = ``new` `int``[N];` `    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``if` `(str.charAt(i) == ``'0'``)``        ``{``            ``Z.add(arr[i]);``        ``}``        ``else``        ``{``            ``O.add(arr[i]);``        ``}``    ``}` `    ``// Sort them in decreasing order``    ``Collections.sort(Z, ``new` `Comparator()``    ``{``        ``public` `int` `compare(Integer a, Integer b)``        ``{``            ``String X = Integer.toString(a);``            ``String Y = Integer.toString(b);``            ` `            ``// Append Y at the end of X``            ``String XY = X + Y;``        ` `            ``// Append X at the end of Y``            ``String YX = Y + X;``        ` `            ``// Compare and return greater``            ``return` `XY.compareTo(YX) > ``0` `? -``1` `: ``1``;``        ``}``    ``});``    ` `    ``Collections.sort(O, ``new` `Comparator()``    ``{``        ``public` `int` `compare(Integer a, Integer b)``        ``{``            ``String X = Integer.toString(a);``            ``String Y = Integer.toString(b);` `            ``// Append Y at the end of X``            ``String XY = X + Y;``        ` `            ``// Append X at the end of Y``            ``String YX = Y + X;``        ` `            ``// Compare and return greater``            ``return` `XY.compareTo(YX) > ``0` `? -``1` `: ``1``;``        ``}``    ``});``    ` `    ``int` `j = ``0``, k = ``0``;` `    ``// Generate the sorted array``    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``if` `(str.charAt(i) == ``'0'``)``        ``{``            ``ans[i] = Z.get(j++);``        ``}``        ``else``        ``{``            ``ans[i] = O.get(k++);``        ``}``    ``}` `    ``for``(``int` `i = ``0``; i < N; i++)``    ``{``        ``System.out.print(ans[i] + ``" "``);``    ``}``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int``[] arr = { ``1``, ``3``, ``456``, ``6``, ``7``, ``8` `};``    ``String str = ``"101101"``;``    ` `    ``findMaxArray(arr, str);``}``}` `// This code is contributed by offbeat`
Output:
```8 7 6 456 3 1

```

Time Complexity: O(NlogN)
Auxiliary Space: O(N)

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