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[0] is less than arr[2], 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:
- Create two arrays to store 0-charactered index elements and 1-charactered index elements from the array.
- Sort the arrays to form largest possible numbers from these two arrays.
- 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 <bits/stdc++.h>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<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 Codeint 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 formedstatic void findMaxArray(int[] arr, String str){ int N = arr.length; ArrayList<Integer> 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<Integer>() { 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<Integer>() { 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 |
Python3
# Python Program to implement# the above approachfrom functools import cmp_to_key# Function to return the rearranged# array in the form of largest# possible number that can be formeddef findMaxArray(arr, strr): N = len(arr) Z = [] O = [] ans = [0]*N for i in range(N): if (strr[i] == '0'): Z.append(arr[i]) else: O.append(arr[i]) # Sort them in decreasing order Z.sort(key=cmp_to_key(lambda x, y: 1 if str(x)+str(y) < str(y)+str(x) else -1)) O.sort(key=cmp_to_key(lambda x, y: 1 if str(x)+str(y) < str(y)+str(x) else -1)) j = 0 k = 0 # Generate the sorted array for i in range(N): if (strr[i] == '0'): ans[i] = Z[j] j += 1 else: ans[i] = O[k] k += 1 for i in range(N): print(ans[i], end=" ")# Driver Codearr = [1, 3, 456, 6, 7, 8]strr = "101101"findMaxArray(arr, strr)# This code is contributed by rj13to. |
Javascript
// Javascript program to implement// the above approachfunction compare( a, b){ let X = a.toString(); let Y = b.toString(); // Append Y at the end of X let XY = X + Y; // Append X at the end of Y let YX = Y + X; // Compare and return greater return XY.localeCompare(YX) > 0 ? -1 : 1;}// Function to return the rearranged// array in the form of largest// possible number that can be formedfunction findMaxArray( arr, str){ let N = arr.length; let Z = []; let O = []; let ans = []; for(let i=0;i<N;i++) { ans.push(0); } for (let i = 0; i < N; i++) { if (str[i] == '0') { Z.push(arr[i]); } else { O.push(arr[i]); } } // Sort them in decreasing order Z.sort(compare); O.sort(compare); let j = 0, k = 0; // Generate the sorted array for(let i = 0; i < N; i++) { if (str[i] == '0') { ans[i] = Z[j++]; } else { ans[i] = O[k++]; } } console.log(ans);}// Driver codelet arr = [ 1, 3, 456, 6, 7, 8 ];let str = "101101"; findMaxArray(arr, str);// This code is contributed by harimahecha |
C#
// Include namespace systemusing System;using System.Collections.Generic;public class GFG{ // Function to return the rearranged // array in the form of largest // possible number that can be formed public static void findMaxArray(int[] arr, String str) { var N = arr.Length; var Z = new List<int>(); var O = new List<int>(); int[] ans = new int[N]; for (int i = 0; i < N; i++) { if (str[i] == '0') { Z.Add(arr[i]); } else { O.Add(arr[i]); } } // Sort them in decreasing order Z.Sort((a,b) =>{ var X = Convert.ToString(a); var Y = Convert.ToString(b); // Append Y at the end of X var XY = X + Y; // Append X at the end of Y var YX = Y + X; // Compare and return greater return string.CompareOrdinal(XY,YX) > 0 ? -1 : 1; }); O.Sort((a,b) =>{ var X = Convert.ToString(a); var Y = Convert.ToString(b); // Append Y at the end of X var XY = X + Y; // Append X at the end of Y var YX = Y + X; // Compare and return greater return string.CompareOrdinal(XY,YX) > 0 ? -1 : 1; }); var j = 0; var 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++) { Console.Write(ans[i].ToString() + " "); } } // Driver code public static void Main(String[] args) { int[] arr = {1, 3, 456, 6, 7, 8}; var str = "101101"; GFG.findMaxArray(arr, str); }} |
Output:
8 7 6 456 3 1
Time Complexity: O(NlogN)
Auxiliary Space: O(N)
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