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 formedvoid findMaxArray(vector& arr, string& str){    int N = arr.size();    vector 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 Codeint main(){    vector arr = { 1, 3, 456, 6, 7, 8 };    string str = "101101";    findMaxArray(arr, str);    return 0;}

Java

 // Java program to implement// the above approachimport java.util.*;import java.lang.*; class GFG{ // Function to return the rearranged// array in the form of largest// possible number that can be formedstatic 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 codepublic 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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