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:
Below is the implementation of the above approach:
// 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 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;
}
Output:
8 7 6 456 3 1
Time Complexity:_ O(NlogN)_
Auxiliary Space:_ O(N)_
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