# Elements of first array that have more frequencies

Given two arrays (which may or may not be sorted). There arrays are such that they might have some common elements in them. We need to find elements whose counts of occurrences are more in first array than second.

Examples:

```Input : ar1[] = {1, 2, 2, 2, 3, 3, 4, 5}
ar2[] = {2, 2, 3, 3, 3, 4}
Output : 1 2 5
1 occurs one times in first and zero times in second
2 occurs three times in first and two times in second
............................

Input : ar1[] = {1, 3, 4, 2, 3}
ar2[] = {3, 4, 5}
Output : 3
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to use hashing. We traverse first array and insert all elements and their frequencies in a hash table. Now we traverse through the second array and reduce frequencies in hash table for the common elements. Now we traverse through first array again and print those elements whose frequencies are still more than 0. To avoid repeated printing of same elements, we set frequency as 0.

## C++

 `// C++ program to print all those elements of ` `// first array that have more frequencies than ` `// second array. ` `#include ` `using` `namespace` `std; ` ` `  `// Compares two intervals according to staring times. ` `void` `moreFreq(``int` `ar1[], ``int` `ar2[], ``int` `m, ``int` `n) ` `{ ` `    ``// Traverse first array and store frequencies ` `    ``// of all elements ` `    ``unordered_map<``int``, ``int``> mp; ` `    ``for` `(``int` `i = 0; i < m; i++) ` `        ``mp[ar1[i]]++; ` ` `  `    ``// Traverse second array and reduce frequencies ` `    ``// of common elements. ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``if` `(mp.find(ar2[i]) != mp.end()) ` `            ``mp[ar2[i]]--; ` ` `  `    ``// Now traverse first array again and print ` `    ``// all those elements whose frequencies are ` `    ``// more than 0. To avoid repeated printing, ` `    ``// we set frequency as 0 after printing. ` `    ``for` `(``int` `i = 0; i < m; i++) { ` `        ``if` `(mp[ar1[i]] > 0) { ` `            ``cout << ar1[i] << ``" "``; ` `            ``mp[ar1[i]] = 0; ` `        ``} ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `ar1[] = { 1, 2, 2, 2, 3, 3, 4, 5 }; ` `    ``int` `ar2[] = { 2, 2, 3, 3, 3, 4 }; ` `    ``int` `m = ``sizeof``(ar1) / ``sizeof``(ar1); ` `    ``int` `n = ``sizeof``(ar2) / ``sizeof``(ar2); ` `    ``moreFreq(ar1, ar2, m, n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to print all those elements of ` `// first array that have more frequencies than ` `// second array. ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `    ``// Compares two intervals according to staring times. ` `    ``static` `void` `moreFreq(``int` `ar1[], ``int` `ar2[], ``int` `m, ``int` `n) ` `    ``{ ` `        ``// Traverse first array and store frequencies ` `        ``// of all elements ` `        ``Map mp = ``new` `HashMap<>(); ` `        ``for` `(``int` `i = ``0` `; i < m; i++) ` `        ``{ ` `            ``if``(mp.containsKey(ar1[i])) ` `            ``{ ` `                ``mp.put(ar1[i], mp.get(ar1[i])+``1``); ` `            ``} ` `            ``else` `            ``{ ` `                ``mp.put(ar1[i], ``1``); ` `            ``} ` `        ``} ` `         `  `        ``// Traverse second array and reduce frequencies ` `        ``// of common elements. ` `        ``for` `(``int` `i = ``0``; i < n; i++) ` `            ``if` `(mp.containsKey(ar2[i])) ` `                ``mp.put(ar2[i], mp.get(ar2[i])-``1``); ` `     `  `        ``// Now traverse first array again and print ` `        ``// all those elements whose frequencies are ` `        ``// more than 0. To avoid repeated printing, ` `        ``// we set frequency as 0 after printing. ` `        ``for` `(``int` `i = ``0``; i < m; i++)  ` `        ``{ ` `            ``if` `(mp.get(ar1[i]) > ``0``)  ` `            ``{ ` `                ``System.out.print(ar1[i] + ``" "``); ` `                ``mp.put(ar1[i], ``0``); ` `            ``} ` `        ``} ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args)  ` `    ``{ ` `        ``int` `ar1[] = { ``1``, ``2``, ``2``, ``2``, ``3``, ``3``, ``4``, ``5` `}; ` `        ``int` `ar2[] = { ``2``, ``2``, ``3``, ``3``, ``3``, ``4` `}; ` `        ``int` `m = ar1.length; ` `        ``int` `n = ar2.length; ` `        ``moreFreq(ar1, ar2, m, n); ` `    ``} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## Python3

 `# Python3 program to print all those elements of ` `# first array that have more frequencies than ` `# second array. ` `import` `math as mt ` ` `  `# Compares two intervals according to ` `# staring times. ` `def` `moreFreq(ar1, ar2, m, n): ` ` `  `    ``# Traverse first array and store  ` `    ``# frequencies of all elements ` `    ``mp ``=` `dict``() ` `    ``for` `i ``in` `range``(m): ` `        ``if` `ar1[i] ``in` `mp.keys(): ` `            ``mp[ar1[i]] ``+``=` `1` `        ``else``: ` `            ``mp[ar1[i]] ``=` `1` `                 `  `    ``# Traverse second array and reduce  ` `    ``# frequencies of common elements. ` `    ``for` `i ``in` `range``(n): ` `        ``if` `ar2[i] ``in` `mp.keys(): ` `            ``mp[ar2[i]] ``-``=` `1` ` `  `    ``# Now traverse first array again and print ` `    ``# all those elements whose frequencies are ` `    ``# more than 0. To avoid repeated printing, ` `    ``# we set frequency as 0 after printing. ` `    ``for` `i ``in` `range``(m): ` `        ``if` `(mp[ar1[i]] > ``0``): ` `            ``print``(ar1[i], end ``=` `" "``) ` `            ``mp[ar1[i]] ``=` `0` `                                 `  `# Driver code ` `ar1 ``=` `[ ``1``, ``2``, ``2``, ``2``, ``3``, ``3``, ``4``, ``5` `] ` `ar2 ``=` `[ ``2``, ``2``, ``3``, ``3``, ``3``, ``4` `] ` `m ``=` `len``(ar1) ` `n ``=` `len``(ar2)  ` `moreFreq(ar1, ar2, m, n) ` ` `  `# This code is contributed  ` `# by mohit kumar 29 `

## C#

 `// C# pprogram to print all those elements of  ` `// first array that have more frequencies than  ` `// second array.  ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG  ` `{  ` ` `  `    ``// Compares two intervals according to ` `    ``// staring times.  ` `    ``static` `void` `moreFreq(``int` `[]ar1, ``int` `[]ar2, ` `                         ``int` `m, ``int` `n)  ` `    ``{  ` `        ``// Traverse first array and store frequencies  ` `        ``// of all elements  ` `        ``Dictionary<``int``,  ` `                   ``int``> mp = ``new` `Dictionary<``int``,  ` `                                            ``int``>();  ` `        ``for` `(``int` `i = 0 ; i < m; i++)  ` `        ``{  ` `            ``if``(mp.ContainsKey(ar1[i]))  ` `            ``{  ` `                ``mp[ar1[i]] = mp[ar1[i]] + 1;  ` `            ``}  ` `            ``else` `            ``{  ` `                ``mp.Add(ar1[i], 1);  ` `            ``}  ` `        ``}  ` `         `  `        ``// Traverse second array and reduce frequencies  ` `        ``// of common elements.  ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``if` `(mp.ContainsKey(ar2[i]))  ` `                ``mp[ar2[i]] = mp[ar2[i]] - 1;  ` `     `  `        ``// Now traverse first array again and print  ` `        ``// all those elements whose frequencies are  ` `        ``// more than 0. To avoid repeated printing,  ` `        ``// we set frequency as 0 after printing.  ` `        ``for` `(``int` `i = 0; i < m; i++)  ` `        ``{  ` `            ``if` `(mp[ar1[i]] > 0)  ` `            ``{  ` `                ``Console.Write(ar1[i] + ``" "``);  ` `                ``mp[ar1[i]] = 0;  ` `            ``}  ` `        ``}  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main(String[] args)  ` `    ``{  ` `        ``int` `[]ar1 = { 1, 2, 2, 2, 3, 3, 4, 5 };  ` `        ``int` `[]ar2 = { 2, 2, 3, 3, 3, 4 };  ` `        ``int` `m = ar1.Length;  ` `        ``int` `n = ar2.Length;  ` `        ``moreFreq(ar1, ar2, m, n);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by 29AjayKumar  `

Output:

```1 2 5
```

Time Complexity : O(m + n) under the assumption that unordered_map find() and insert() work in O(1) time.

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