Given two unsorted arrays arr1[] and arr2[]. They may contain duplicates. For each element in arr1[] count elements less than or equal to it in array arr2[].
Examples:
Input : arr1[] = [1, 2, 3, 4, 7, 9]
arr2[] = [0, 1, 2, 1, 1, 4]
Output : [4, 5, 5, 6, 6, 6]
Explanation: There are 4 elements less than or equal to 1 in second array, similarly there are 5 elements less than 2 in second array, calculate the values similarly for other elements.Input : arr1[] = [5, 10, 2, 6, 1, 8, 6, 12]
arr2[] = [6, 5, 11, 4, 2, 3, 7]
Output : [4, 6, 1, 5, 0, 6, 5, 7]
Explanation: There are 4 elements less than or equal to 5 in second array, similarly there are 6 elements less than 10 in second array, calculate the values similarly for other elements.
This problem is already discussed in the previous post.
Solution: In this article, a more optimized linear time solution to the above problem is discussed. The approach discussed here works for arrays with values in a small range. A range of values that can be used as an index in an array.
Approach: The idea is to create a prefix map up to the maximum element of the second array. The prefix array will store the maximum element up to that index, for example, prefix[i] will store the count of elements up to i. Then traverse through the first array and find the count of elements less than or equal to that element from the prefix array.
The prefix array is created by traversing through the prefix array and updating the current element by adding the precious element, i.e. prefix[i]+ =prefix[i-1].
Algorithm:
- Create extra space (prefix) of the size of a maximum element of the second and first array or a map.
- Traverse the second array.
- For every element of the second array increase the count of prefix array, i.e. prefix[arr2[i]]++
- Traverse through the prefix array from 1 to MAX (maximum element of the second and first array and update the ith element by adding the sum of i-1th element
- Now traverse through the first array and print the value of prefix[arr_1[i]]
Implementation:
C++
// C++ program for each element in 1st // array count elements less than or equal to it // in 2nd array #include <iostream> using namespace std; #define MAX 100000 // Function for each element in 1st // array count elements less than or equal to it // in 2nd array void countEleLessThanOrEqual(int arr1[], int m, int arr2[], int n) { // Creating hash array initially // filled with zero int hash[MAX] = {0}; // Insert element of arr2[] to hash // such that hash[i] will give count of // element i in arr2[] for (int i = 0; i < n; i++) hash[arr2[i]]++; // Presum of hash array // such that hash[i] will give count of // element less than or equals to i in arr2[] for (int i=1; i<MAX; i++) hash[i] = hash[i] + hash[i-1]; // Traverse arr1[] and print hash[arr[i]] for (int i = 0; i < m; i++) cout << hash[arr1[i]] << " "; } // Driver code int main() { int arr1[] = {1, 2, 3, 4, 7, 9}; int arr2[] = {0, 1, 2, 1, 1, 4}; int m, n; m = sizeof(arr1) / sizeof(arr1[0]); n = sizeof(arr2) / sizeof(arr2[0]); countEleLessThanOrEqual(arr1, m, arr2, n); return 0; } |
Java
// Java program for each element // in 1st array count elements // less than or equal to it in // 2nd array import java.io.*;class GFG {static int MAX = 100000;// Function for each element // in 1st array count elements // less than or equal to it // in 2nd array static void countEleLessThanOrEqual(int arr1[], int m, int arr2[], int n){ // Creating hash array initially // filled with zero int hash[] = new int[MAX]; // Insert element of arr2[] to // hash such that hash[i] will // give count of element i in arr2[] for (int i = 0; i < n; i++) hash[arr2[i]]++; // Presum of hash array // such that hash[i] will // give count of element // less than or equals to // i in arr2[] for(int i = 1; i < MAX; i++) { hash[i] = hash[i] + hash[i - 1]; } // Traverse arr1[] and // print hash[arr[i]] for (int i = 0; i < m; i++) { System.out.print(hash[arr1[i]] + " "); }}// Driver codepublic static void main (String[] args) { int arr1[] = {1, 2, 3, 4, 7, 9}; int arr2[] = {0, 1, 2, 1, 1, 4}; int m, n; m = arr1.length; n = arr2.length; countEleLessThanOrEqual(arr1, m, arr2, n);}}// This code is contributed// by inder_verma |
Python3
# Python 3 program for each element in 1st # array count elements less than or equal # to it in 2nd array MAX = 100000# Function for each element in 1st # array count elements less than or # equal to it in 2nd array def countEleLessThanOrEqual(arr1, m, arr2, n): # Creating hash array initially # filled with zero hash = [0 for i in range(MAX)] # Insert element of arr2[] to hash # such that hash[i] will give count # of element i in arr2[] for i in range(n): hash[arr2[i]] += 1 # Presum of hash array such that # hash[i] will give count of element # less than or equals to i in arr2[] for i in range(1, MAX, 1): hash[i] = hash[i] + hash[i - 1] # Traverse arr1[] and print hash[arr[i]] for i in range(m): print(hash[arr1[i]], end = " ") # Driver code if __name__ == '__main__': arr1 = [1, 2, 3, 4, 7, 9] arr2 = [0, 1, 2, 1, 1, 4] m = len(arr1) n = len(arr2) countEleLessThanOrEqual(arr1, m, arr2, n) # This code is contributed by# Shashank_Sharma |
C#
// C# program for each element // in 1st array count elements // less than or equal to it in // 2nd array using System;public class GFG { static int MAX = 100000; // Function for each element // in 1st array count elements // less than or equal to it // in 2nd array static void countEleLessThanOrEqual(int []arr1, int m, int []arr2, int n) { // Creating hash array initially // filled with zero int []hash = new int[MAX]; // Insert element of arr2[] to // hash such that hash[i] will // give count of element i in arr2[] for (int i = 0; i < n; i++) hash[arr2[i]]++; // Presum of hash array // such that hash[i] will // give count of element // less than or equals to // i in arr2[] for(int i = 1; i < MAX; i++) { hash[i] = hash[i] + hash[i - 1]; } // Traverse arr1[] and // print hash[arr[i]] for (int i = 0; i < m; i++) { Console.Write(hash[arr1[i]] + " "); } } // Driver code public static void Main () { int []arr1 = {1, 2, 3, 4, 7, 9}; int []arr2 = {0, 1, 2, 1, 1, 4}; int m, n; m = arr1.Length; n = arr2.Length; countEleLessThanOrEqual(arr1, m, arr2, n); } } // This code is contributed by Shikha Singh. |
PHP
<?php // PHP program for each element in 1st // array count elements less than or // equal to it in 2nd array $MAX = 100000 ;// Function for each element in 1st // array count elements less than or // equal to it in 2nd array function countEleLessThanOrEqual(&$arr1, $m, &$arr2, $n) { global $MAX; // Creating hash array initially // filled with zero $hash = array_fill(0, $MAX, NULL); // Insert element of arr2[] to hash // such that hash[i] will give count of // element i in arr2[] for ($i = 0; $i < $n; $i++) $hash[$arr2[$i]]++; // Presum of hash array such that hash[i] // will give count of element less than // or equals to i in arr2[] for ($i = 1; $i < $MAX; $i++) $hash[$i] = $hash[$i] + $hash[$i - 1]; // Traverse arr1[] and print hash[arr[i]] for ($i = 0; $i < $m; $i++) echo $hash[$arr1[$i]] . " "; } // Driver code $arr1 = array(1, 2, 3, 4, 7, 9);$arr2 = array(0, 1, 2, 1, 1, 4); $m = sizeof($arr1); $n = sizeof($arr2); countEleLessThanOrEqual($arr1, $m, $arr2, $n); // This code is contributed // by ChitraNayal?> |
Javascript
<script>// JavaScript program for each element// in 1st array count elements// less than or equal to it in// 2nd array let MAX = 100000; // Function for each element// in 1st array count elements// less than or equal to it// in 2nd array function countEleLessThanOrEqual(arr1,m,arr2,n) { // Creating hash array initially // filled with zero let hash = new Array(MAX); for(let i=0;i<hash.length;i++) { hash[i]=0; } // Insert element of arr2[] to // hash such that hash[i] will // give count of element i in arr2[] for (let i = 0; i < n; i++) hash[arr2[i]]++; // Presum of hash array // such that hash[i] will // give count of element // less than or equals to // i in arr2[] for(let i = 1; i < MAX; i++) { hash[i] = hash[i] + hash[i - 1]; } // Traverse arr1[] and // print hash[arr[i]] for (let i = 0; i < m; i++) { document.write(hash[arr1[i]] + " "); } } // Driver code let arr1=[1, 2, 3, 4, 7, 9]; let arr2=[0, 1, 2, 1, 1, 4]; let m, n; m = arr1.length; n = arr2.length; countEleLessThanOrEqual(arr1, m, arr2, n); // This code is contributed by avanitrachhadiya2155</script> |
Output
4 5 5 6 6 6
Complexity Analysis:
- Time Complexity: O(max), where max is the maximum element of both arrays
- Auxiliary Space: O(max), where max is the maximum element of both arrays.
Implementation:
C++
// C++ program for each element in 1st // array count elements less than or equal to it // in 2nd array #include <iostream>#include <map>#include <vector>// Function for each element in 1st // array count elements less than or equal to it // in 2nd array void countLessThanOrEqual(const std::vector<int>& vec1, const std::vector<int>& vec2) { std::map<int, unsigned int> countOfVec2; for (const auto& item : vec2) { ++countOfVec2[item]; } unsigned int prev = 0; for (auto& pair : countOfVec2) { pair.second += prev; prev = pair.second; } // Traverse arr1[] and print result for (const auto& item : vec1) { unsigned int result = (--countOfVec2.upper_bound(item))->second; std::cout << result << " "; }}// Driver codeint main(){ std::vector<int> arr1 = { 1, 2, 3, 4, 7, 9 }; std::vector<int> arr2 = { 0, 1, 2, 1, 1, 4 }; countLessThanOrEqual(arr1, arr2); return 0;} |
Java
// Java program for each element in 1st // array count elements less than or equal to it // in 2nd arrayimport java.util.*; class GFG { // Function for each element in 1st // array count elements less than or equal to it // in 2nd array static void countLessThanOrEqual(int vec1[], int vec2[]) { HashMap<Integer, Integer> countOfVec2 = new HashMap<>(); for(int item : vec2) { if(!countOfVec2.containsKey(item)) countOfVec2.put(item, 0); countOfVec2.put(item, countOfVec2.get(item) + 1); } int prev = 0; Vector<Integer> li = new Vector<Integer>(); for (Map.Entry<Integer, Integer> pair : countOfVec2.entrySet()) { li.add(pair.getKey()); } for(int pair : li) { countOfVec2.put(pair, countOfVec2.get(pair) + prev); prev = countOfVec2.get(pair); } // Traverse arr1[] and print result for(int item : vec1) { int i = 0, v = 0, last = 0; for (Map.Entry<Integer, Integer> pair : countOfVec2.entrySet()) { last = pair.getKey(); if(item < pair.getKey()) { v = i; break; } i++; } v -= 1; if(v == -1) { v = last; } int result = countOfVec2.get(v); System.out.print(result + " "); } } // Driver code public static void main(String[] args) { int arr1[] = {1, 2, 3, 4, 7, 9}; int arr2[] = {0, 1, 2, 1, 1, 4}; countLessThanOrEqual(arr1, arr2); }}// This code is contributed by divyesh072019. |
Python3
# Python3 program for each element in 1st # array count elements less than or equal to it # in 2nd array # Function for each element in 1st # array count elements less than or equal to it # in 2nd array def countLessThanOrEqual(vec1, vec2): countOfVec2 = {} for item in vec2: if item not in countOfVec2: countOfVec2[item] = 0 countOfVec2[item] += 1 prev = 0 for pair in countOfVec2: countOfVec2[pair] += prev prev = countOfVec2[pair] val = list(countOfVec2) # Traverse arr1[] and print result for item in vec1: i = 0 v = 0 for i in range(len(val)): if item < val[i]: v = i break v -= 1 if v == -1: v = val[-1] result = countOfVec2[v] print(result, end = " ")# Driver codearr1 = [1, 2, 3, 4, 7, 9]arr2 = [0, 1, 2, 1, 1, 4]countLessThanOrEqual(arr1, arr2)# This code is contributed by shubhamsingh10 |
C#
// C# program for each element in 1st // array count elements less than or equal to it // in 2nd array using System;using System.Collections.Generic; class GFG { // Function for each element in 1st // array count elements less than or equal to it // in 2nd array static void countLessThanOrEqual(int[] vec1, int[] vec2) { Dictionary<int, int> countOfVec2 = new Dictionary<int, int>(); foreach(int item in vec2) { if(!countOfVec2.ContainsKey(item)) countOfVec2[item] = 0; countOfVec2[item] += 1; } int prev = 0; List<int> li = new List<int>(); foreach(KeyValuePair<int, int> pair in countOfVec2) { li.Add(pair.Key); } foreach(int pair in li) { countOfVec2[pair] += prev; prev = countOfVec2[pair]; } // Traverse arr1[] and print result foreach(int item in vec1) { int i = 0, v = 0, last = 0; foreach(KeyValuePair<int, int> pair in countOfVec2) { last = pair.Key; if(item < pair.Key) { v = i; break; } i++; } v -= 1; if(v == -1) { v = last; } int result = countOfVec2[v]; Console.Write(result + " "); } } // Driver code static void Main() { int[] arr1 = {1, 2, 3, 4, 7, 9}; int[] arr2 = {0, 1, 2, 1, 1, 4}; countLessThanOrEqual(arr1, arr2); }}// This code is contributed by divyeshrabadiya07. |
Javascript
<script>// Javascript program for each element in 1st // array count elements less than or equal to it // in 2nd array // Function for each element in 1st // array count elements less than or equal to it // in 2nd array function countLessThanOrEqual(vec1,vec2) { let countOfVec2 = new Map(); for(let item = 0;item < vec2.length; item++) { if(!countOfVec2.has(vec2[item])) countOfVec2.set(vec2[item], 0); countOfVec2.set(vec2[item], countOfVec2.get(vec2[item]) + 1); } let prev = 0; let li = []; for (let [key, value] of countOfVec2.entries()) { li.push(key); } for(let pair = 0; pair < li.length; pair++) { countOfVec2.set(pair, countOfVec2.get(li[pair]) + prev); prev = countOfVec2.get(li[pair]); } // Traverse arr1[] and print result for(let item=0; item<vec1.length;item++) { let i = 0, v = 0, last = 0; for (let [key, value] of countOfVec2.entries()) { last = key; if(vec1[item] < key) { v = i; break; } i++; } v -= 1; if(v == -1) { v = last; } let result = countOfVec2.get(v); document.write(result + " "); } } // Driver code let arr1=[1, 2, 3, 4, 7, 9]; let arr2=[0, 1, 2, 1, 1, 4]; countLessThanOrEqual(arr1, arr2); // This code is contributed by rag2127</script> |
Output
4 5 5 6 6 6
Complexity Analysis:
- Time Complexity: O(max) where max is the maximum element of both arrays.
- Auxiliary Complexity : O(max) where max is the maximum element of both arrays.
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