Given two arrays A and B of the same size N. Check whether array A can be fit into array B. An array is said to fit into another array if by arranging the elements of both arrays, there exists a solution such that the ith element of the first array is less than or equal to ith element of the second array.
Examples:Â
Input : A[] = { 7, 5, 3, 2 }, B[] = { 5, 4, 8, 7 }
Output : YES
Rearrange the first array to {3, 2, 7, 5}
Do not rearrange the second array's element.
After rearranging, all Ai<=Bi.
Input : A[] = { 7, 5, 3, 2, 5, 105, 45, 10 }, B[] = { 2, 4, 0, 5, 6, 9, 75, 84 }
Output : NO
Approach: Sort both the arrays and check whether Ai is less than or equal to Bi for all 0 ≤ i ≤ N. If at any ith position Ai is greater than Bi return false, otherwise return true.Â
Steps to solve the problem:
- Sort array A in non-decreasing order.
- Sort array B in non-decreasing order.
- Â For each element i from 0 to N-1 do the following:
- Â If A[i] > B[i], return false.
- If the loop completes without returning false, return true.
Below is the implementation of the above approach:Â
C++
// C++ Program to check whether an array// can be fit into another array with given// condition.#include <bits/stdc++.h>using namespace std;Â
// Returns true if the array A can be fit into// array B, otherwise falsebool checkFittingArrays(int A[], int B[], int N){    // Sort both the arrays    sort(A, A + N);    sort(B, B + N);Â
    // Iterate over the loop and check whether every    // array element of A is less than or equal to    // its corresponding array element of B    for (int i = 0; i < N; i++)        if (A[i] > B[i])            return false;Â
    return true;}Â
// Driver Codeint main(){Â Â Â Â int A[] = { 7, 5, 3, 2 };Â Â Â Â int B[] = { 5, 4, 8, 7 };Â Â Â Â int N = sizeof(A) / sizeof(A[0]);Â
    if (checkFittingArrays(A, B, N))        cout << "YES";    else        cout << "NO";    return 0;} |
Java
// Java Program to check// whether an array can// be fit into another// array with given// condition.import java.io.*;import java.util.*;import java.lang.*;Â
class GFG {Â
// Returns true if the// array A can be fit // into array B, // otherwise falsestatic boolean checkFittingArrays(int []A,                                  int []B,                                   int N){         // Sort both the arrays    Arrays.sort(A);    Arrays.sort(B);Â
    // Iterate over the loop    // and check whether every    // array element of A is     // less than or equal to    // its corresponding array     // element of B    for (int i = 0; i < N; i++)        if (A[i] > B[i])            return false;Â
    return true;}Â
// Driver Codepublic static void main(String[] args){    int A[] = {7, 5, 3, 2};    int B[] = {5, 4, 8, 7};    int N = A.length;         if (checkFittingArrays(A, B, N))        System.out.print("YES");    else        System.out.print("NO");}} |
Python3
# Python3 Program to check whether an array# can be fit into another array with given# condition.Â
# Returns true if the array A can be fit into# array B, otherwise falsedef checkFittingArrays(A, B, N):         # Sort both the arrays    A = sorted(A)    B = sorted(B)Â
    # Iterate over the loop and check whether     # every array element of A is less than     # or equal to its corresponding array     # element of B    for i in range(N):        if (A[i] > B[i]):            return FalseÂ
    return TrueÂ
# Driver CodeA = [7, 5, 3, 2]B = [5, 4, 8, 7]N = len(A)Â
if (checkFittingArrays(A, B, N)):Â Â Â Â print("YES")else:Â Â Â Â print("NO")Â
# This code is contributed # by mohit kumar |
C#
// C# Program to check// whether an array can// be fit into another// array with given// condition.using System;Â
class GFG {Â
// Returns true if the// array A can be fit // into array B, // otherwise falsestatic bool checkFittingArrays(int []A,                               int []B,                                int N){         // Sort both the arrays    Array.Sort(A);    Array.Sort(B);Â
    // Iterate over the loop    // and check whether every    // array element of A is     // less than or equal to    // its corresponding array     // element of B    for (int i = 0; i < N; i++)        if (A[i] > B[i])            return false;Â
    return true;}Â
// Driver Codepublic static void Main () {    int []A = {7, 5, 3, 2};    int []B = {5, 4, 8, 7};    int N = A.Length;         if (checkFittingArrays(A, B, N))        Console.WriteLine("YES");    else        Console.WriteLine("NO");}}Â
// This code is contributed // by anuj_67. |
PHP
<?php// PHP Program to check whether an // array can be fit into another // array with given condition.// Returns true if the array A can // be fit into array B, otherwise falsefunction checkFittingArrays($A, $B, $N){    // Sort both the arrays    sort($A);    sort($B);Â
    // Iterate over the loop and check     // whether every array element of     // A is less than or equal to its    // corresponding array element of B    for ($i = 0; $i < $N; $i++)        if ($A[$i] > $B[$i])            return false;Â
    return true;}Â
// Driver Code$A = array( 7, 5, 3, 2 );$B = array( 5, 4, 8, 7 );$N = count($A);Â
if (checkFittingArrays($A, $B, $N))    echo "YES";else    echo "NO";     // This code is contributed by shs?> |
Javascript
<script>// Javascript Program to check// whether an array can// be fit into another// array with given// condition.         // Returns true if the// array A can be fit // into array B, // otherwise false    function checkFittingArrays(A, B, N)    {             // Sort both the arrays    A.sort(function(a, b){return a - b;});    B.sort(function(a, b){return a - b;});       // Iterate over the loop    // and check whether every    // array element of A is     // less than or equal to    // its corresponding array     // element of B    for (let i = 0; i < N; i++)        if (A[i] > B[i])            return false;       return true;    }         // Driver Code    let A = [7, 5, 3, 2];    let B = [5, 4, 8, 7];    let N = A.length;    if (checkFittingArrays(A, B, N))        document.write("YES");    else        document.write("NO");       // This code is contributed by unknown2108</script> |
Output
YES
Time Complexity: O(N * logN), where N is the size of the array.
Auxiliary Space: O(1)
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