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Check if mapping is possible to make sum of first array larger than second array

  • Last Updated : 07 Jul, 2021

Given a positive integer K and two arrays A[] and B[] of size N and M respectively, each containing elements in the range [1, K]. The task is to map all numbers from 1 to K with a function f, such that f(1)<f(2)<f(3)<…<f(K), and find whether the sum of f(A[i]) for 0≤i<N is greater than the sum of f(B[i]) for 0≤i<M.

Examples:

Input: A[] = {2, 2, 2}, B[] = {1, 1, 3}, K=3
Output: YES
Explanation: One possible way is to assign the following values to the function f as follows: 
f(1)=1
f(2)=2
f(3)=3
Sum of f(A[i]) for every i = 2+2+2 = 6 and sum of f(B[i]) for every i = 1+1+3 = 5.

Input: A[] = {8, 2, 8, 6, 9, 10}, B[] = {1, 2, 3, 4}, K=10
Output: YES

Approach: Follow the steps below to solve the problem:



  • If the value of N>M, the print “YES” as the answer, since an assignment can always be made such that the sum of A is greater than B.
  • Otherwise, sort the arrays A[] and B[] in descending order.
  • Iterate in the range [0, N-1], using the variable i and if for any index, A[i]>B[i], print “YES” as the answer since assigning a large enough value to A[i] will make the sum of A[] is greater than B[].
  • Otherwise, print “NO” as the result.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if mapping is
// possible to make sum of first array
// larger than second array
int isPossible(int A[], int B[], int K, int N, int M)
{
    // If N > M, an assignment is
    // always possible
    if (N > M)
        return 1;
 
    // Sort A[] in descending order
    sort(A, A + N, greater<int>());
 
    // Sort B[] in descending order
    sort(B, B + M, greater<int>());
 
    // Traverse in the arrays
    for (int i = 0; i < N; i++)
 
        // If A[i] > B[i], assignment
        // is possible
        if (A[i] > B[i])
            return 1;
   
    return 0;
}
 
// Driver Code
int main()
{
    // Given Input
    int A[] = { 2, 2, 2 };
    int B[] = { 1, 1, 3 };
    int K = 3;
    int N = sizeof(A) / sizeof(A[0]);
    int M = sizeof(B) / sizeof(B[0]);
 
    // Function Call
    if (isPossible(A, B, K, N, M))
        cout << "YES";
    else
        cout << "NO";
 
    return 0;
}

Java




// Java Program for the above approach
import java.io.*;
import java.util.Arrays;
import java.util.Collections;
 
class GFG {
    public static void reverse(int array[])
    {
 
        // Length of the array
        int n = array.length;
 
        // Swaping the first half elements with last half
        // elements
        for (int i = 0; i < n / 2; i++) {
 
            // Storing the first half elements temporarily
            int temp = array[i];
 
            // Assigning the first half to the last half
            array[i] = array[n - i - 1];
 
            // Assigning the last half to the first half
            array[n - i - 1] = temp;
        }
    }
    // Function to check if mapping is
    // possible to make sum of first array
    // larger than second array
    public static int isPossible(int A[], int B[], int K,
                                 int N, int M)
    {
        // If N > M, an assignment is
        // always possible
        if (N > M)
            return 1;
 
        // Sort A[] in descending order
        Arrays.sort(A);
        reverse(A);
 
        // Sort B[] in descending order
        Arrays.sort(B);
        reverse(B);
 
        // Traverse in the arrays
        for (int i = 0; i < N; i++)
 
            // If A[i] > B[i], assignment
            // is possible
            if (A[i] > B[i])
                return 1;
 
        return 0;
    }
 
    public static void main(String[] args)
    {
        int A[] = { 2, 2, 2 };
        int B[] = { 1, 1, 3 };
        int K = 3;
        int N = A.length;
        int M = B.length;
 
        // Function Call
        if (isPossible(A, B, K, N, M) == 1)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}
 
// This code is contributed by lokeshpotta20

Python3




# Python3 program for the above approach
 
# Function to check if mapping is
# possible to make sum of first array
# larger than second array
def isPossible(A, B, K, N, M):
     
    # If N > M, an assignment is
    # always possible
    if (N > M):
        return 1
 
    # Sort A[] in descending order
    A.sort(reverse = True)
 
    # Sort B[] in descending order
    B.sort(reverse = True)
 
    # Traverse in the arrays
    for i in range(N):
 
        # If A[i] > B[i], assignment
        # is possible
        if (A[i] > B[i]):
            return 1
   
    return 0
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    A = [ 2, 2, 2 ]
    B = [ 1, 1, 3 ]
    K = 3
    N = len(A)
    M = len(B)
 
    # Function Call
    if (isPossible(A, B, K, N, M)):
        print("YES")
    else:
        print("NO")
     
# This code is contributed by SURENDRA_GANGWAR

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to check if mapping is
// possible to make sum of first array
// larger than second array
static int isPossible(int []A, int []B, int K,
                      int N, int M)
{
     
    // If N > M, an assignment is
    // always possible
    if (N > M)
        return 1;
 
    // Sort A[] in descending order
    Array.Sort(A);
    Array.Reverse(A);
 
    // Sort B[] in descending order
    Array.Sort(B);
    Array.Reverse(B);
 
    // Traverse in the arrays
    for(int i = 0; i < N; i++)
     
        // If A[i] > B[i], assignment
        // is possible
        if (A[i] > B[i])
            return 1;
   
    return 0;
}
 
// Driver Code
public static void Main()
{
     
    // Given Input
    int []A = { 2, 2, 2 };
    int []B = { 1, 1, 3 };
    int K = 3;
    int N = A.Length;
    int M = B.Length;
 
    // Function Call
    if (isPossible(A, B, K, N, M) == 1)
        Console.Write("YES");
    else
        Console.Write("NO");
}
}
 
// This code is contributed by ipg2016107

Javascript




<script>
        // JavaScript program for the above approach
 
        // Function to check if mapping is
        // possible to make sum of first array
        // larger than second array
        function isPossible(A, B, K, N, M) {
            // If N > M, an assignment is
            // always possible
            if (N > M)
                return 1;
 
            // Sort A[] in descending order
            A.sort(function (a,b) {return b - a })
 
            // Sort B[] in descending order
            B.sort(function (a,b) {return b - a })
            // Traverse in the arrays
            for (let i = 0; i < N; i++)
 
                // If A[i] > B[i], assignment
                // is possible
                if (A[i] > B[i])
                    return 1;
 
            return 0;
        }
 
        // Driver Code
 
        // Given Input
        var A = [2, 2, 2];
        var B = [1, 1, 3];
        var K = 3;
        var N = A.length;
        var M = B.length;
 
        // Function Call
        if (isPossible(A, B, K, N, M))
            document.write("YES");
        else
            document.write("NO");
 
 
    </script>
Output
YES

Time Complexity: O(N*log(N))
Auxiliary Space: O(1)

 




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