Given an array **arr1** containing integers from **1..N** and another array **arr2** containing integers from **1..M**. Each element of **arr1** is mapped to more than one element in **arr2**. The task is to maximize the number of one to one mappings possible between both arrays.

Here one to one mappings means a unique element of **arr1** must be mapped to unique element of **arr2**.

Mapping is given in the form of vector of pairs **V** where **V.first()** denotes elements of **arr1** and **V.second()** denotes elements of **arr2**.

**Examples:**

Input:N = 3, M = 3 arr1[] = {1, 2, 3} arr2[] = {1, 2, 3} V = { {1, 1}, {2, 1}, {2, 2}, {3, 1}, {3, 2}, {3, 3} }Output:3Explanation:If we carefully see here mappings are 1 --------> {1} 2 --------> {1, 2} 3 --------> {1, 2, 3} so, maximum number of unique pairings can be done as 1 --------> {1} 2 --------> {2} 3 --------> {3}Input:N = 3, M = 1 V = { {1, 1}, {2, 1}, {3, 1} };Output:1Explanation:Either 1 can be mapped or 2 or 3. so maximum one mapping is possible.

**Approach:** The **approach** used here is Greedy.**Preprocessing step**

We are given a vector of pairs which contains mapping information in an unsorted way for eg-

so, we convert this mapping information into a vector of sets **mapping**.

For eg:

This is representing the information in one to many mapping form.

This can be easily understood by the below diagram

**Steps**:

- Find the element in arr1[] which has least number of mappings with elements of arr2[].
- Map this element to first element of its corresponding set.
- Delete all the mappings of this element.
- Also delete all the mappings which contain element from arr2[] those are already mapped.
- Repeat these four steps for all the elements of arr1[].

Below is the implementation of Above approach:

`// C++ Program to convert many to many mapping` `// to maximum no of one to one mappings` ` ` `#include <bits/stdc++.h>` `using` `namespace` `std;` ` ` `// Returns the maximum number` `// of one to one mappings between two arrays` `int` `FindMax(vector<pair<` `int` `, ` `int` `> >& V, ` `int` `N, ` `int` `M)` `{` ` ` `int` `ans = 0;` ` ` ` ` `// Stores whether an element` ` ` `// from the first array is used in mapping` ` ` `bool` `* is_paired = ` `new` `bool` `[N];` ` ` ` ` `// Contains mapping in sorted form` ` ` `vector<set<` `int` `> > mapping(N + 1);` ` ` ` ` `// Initialize all the elements` ` ` `// of first array unused` ` ` `memset` `(is_paired, ` `sizeof` `(is_paired), ` `false` `);` ` ` ` ` `// Insert mappings in sorted form` ` ` `for` `(` `int` `i = 0; i < V.size(); i++) {` ` ` `mapping[V[i].first].insert(V[i].second);` ` ` `}` ` ` ` ` `// While there is always at least one element` ` ` `// which can be mapped to a unique element` ` ` `while` `(` `true` `) {` ` ` `int` `lowest = -1;` ` ` ` ` `// Finds element to be mapped` ` ` `for` `(` `int` `i = 1; i <= N; i++) {` ` ` ` ` `// There is mapping possible` ` ` `if` `(mapping[i].size() > 0) {` ` ` `if` `(lowest == -1)` ` ` `lowest = i;` ` ` `else` `if` `(mapping[i].size()` ` ` `< mapping[lowest].size())` ` ` `lowest = i;` ` ` `}` ` ` `}` ` ` ` ` `if` `(lowest == -1)` ` ` `break` `;` ` ` ` ` `// Element used in the mapping` ` ` `is_paired[lowest] = ` `true` `;` ` ` ` ` `int` `remove` `= *mapping[lowest].begin();` ` ` ` ` `// Delete all the mappings of used element` ` ` `for` `(` `int` `i = 1; i <= N; i++) {` ` ` ` ` `mapping[i].erase(` `remove` `);` ` ` `if` `(i == lowest) {` ` ` `mapping[i].clear();` ` ` `}` ` ` `}` ` ` `ans++;` ` ` `}` ` ` `return` `ans;` `}` ` ` `// Main Function` `int` `main()` `{` ` ` `int` `N = 3;` ` ` `int` `M = 3;` ` ` `int` `arr1[] = { 1, 2, 3 };` ` ` `int` `arr2[] = { 1, 2, 3 };` ` ` ` ` `vector<pair<` `int` `, ` `int` `> >` ` ` `V{ { 1, 1 }, { 2, 1 }, { 2, 2 }, { 3, 1 }, { 3, 2 }, { 3, 3 } };` ` ` ` ` `cout << FindMax(V, N, M) << endl;` ` ` ` ` `return` `0;` `}` |

**Output:**

3

**Time Complexity: ** where x is maximum no one to one mappings.

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