Maximum possible number with concatenations of K numbers from given array
Given an array arr[] of N integers and a positive integer K, the task is to choose K integers from the array arr[] such that concatenation of them forms the largest possible integer. All the array elements must be chosen at least once for creating the number.
Note: It is always guaranteed that N is greater than or equal to K.
Examples:
Input: arr[] = {3, 2, 7}, K = 3
Output: 732
Explanation:
Since each array element has to be used at least once, the biggest possible number is 732.Input: arr[] = {1, 10, 100}, K = 4
Output: 110100100
Approach: The above problem can be solved by sorting and converting numbers to strings. The optimal approach is to take all numbers once. After that, take the number with the most digits. In case of a tie, take the lexicographically largest number. Convert all the numbers in the string using the to_string() function. Follow the below steps to solve the problem:
- Initialize an empty string res to store the answer.
- Sort the array numbers[] in ascending order.
- Initialize a vector of strings v[].
- Iterate over the range [0, K) using the variable i and push the string number numbers[i] into the vector v[].
- Decrease the value of N by K and push the value numbers[K-1] N times into the vector v[].
- Sort the vector v[] using the comparator function.
- Iterate over the range [v.size()-1, 0] using the variable i and add the value v[i] to the variable res.
- After performing the above steps, print the value of res as the answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Custom comparator functionbool str_cmp(string s1, string s2){ return (s1 + s2 < s2 + s1);}// Function to get the largest possible// stringstring largestNumber(vector<int> arr, int K){ int N = arr.size(); // Initialize a new variable which // will store the answers. string res = ""; // Sort the numbers array in // non-decreasing order sort(arr.begin(), arr.end()); // Stores the array element which will // be used to construct the answer vector<string> v; // Take all numbers atleast once for (int i = 0; i < N; i++) { v.push_back(to_string(arr[i])); } K -= N; // Take the largest digits number // for remaining required numbers while (K) { v.push_back(to_string(arr[N - 1])); K--; } // Sort the final answer according to // the comparator function. sort(v.begin(), v.end(), str_cmp); for (int i = v.size() - 1; i >= 0; i--) res += v[i]; return res;}// Driver Codeint main(){ vector<int> arr = { 1, 10, 100 }; int K = 4; cout << largestNumber(arr, K); return 0;} |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to get the largest possible // string function largestNumber(arr, K) { let N = arr.length; // Initialize a new variable which // will store the answers. let res = ""; // Sort the numbers array in // non-decreasing order arr.sort(function (a, b) { return a - b }) // Stores the array element which will // be used to construct the answer let v = []; // Take all numbers atleast once for (let i = 0; i < N; i++) { v.push(arr[i].toString()); } K -= N; // Take the largest digits number // for remaining required numbers while (K) { v.push(arr[N - 1].toString()); K--; } // Sort the final answer according to // the comparator function. v.sort(function (s1, s2) { return parseInt(s1 + s2) - parseInt(s2 + s1) }); for (let i = v.length - 1; i >= 0; i--) res += v[i]; return res; } // Driver Code let arr = [1, 10, 100]; let K = 4; document.write(largestNumber(arr, K)); // This code is contributed by Potta Lokesh </script> |
110100100
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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