A matching in a Bipartite Graph is a set of the edges chosen in such a way that no two edges share an endpoint. A maximum matching is a matching of maximum size (maximum number of edges). In a maximum matching, if any edge is added to it, it is no longer a matching. There can be more than one maximum matchings for a given Bipartite Graph.

**Why do we care?**

There are many real world problems that can be formed as Bipartite Matching. For example, consider the following problem:

*There are M job applicants and N jobs. Each applicant has a subset of jobs that he/she is interested in. Each job opening can only accept one applicant and a job applicant can be appointed for only one job. Find an assignment of jobs to applicants in such that as many applicants as possible get jobs.*

We strongly recommend to read the following post first.

Ford-Fulkerson Algorithm for Maximum Flow Problem

**Maximum Bipartite Matching and Max Flow Problem**

**M**aximum **B**ipartite **M**atching (**MBP**) problem can be solved by converting it into a flow network (See this video to know how did we arrive this conclusion). Following are the steps.

**1) Build a Flow Network**

There must be a source and sink in a flow network. So we add a source and add edges from source to all applicants. Similarly, add edges from all jobs to sink. The capacity of every edge is marked as 1 unit.

2) Find the maximum flow.

We use Ford-Fulkerson algorithm to find the maximum flow in the flow network built in step 1. The maximum flow is actually the MBP we are looking for.

**How to implement the above approach?**

Let us first define input and output forms. Input is in the form of Edmonds matrix which is a 2D array ‘bpGraph[M][N]’ with M rows (for M job applicants) and N columns (for N jobs). The value bpGraph[i][j] is 1 if i’th applicant is interested in j’th job, otherwise 0.

Output is number maximum number of people that can get jobs.

A simple way to implement this is to create a matrix that represents adjacency matrix representation of a directed graph with M+N+2 vertices. Call the fordFulkerson() for the matrix. This implementation requires O((M+N)*(M+N)) extra space.

Extra space can be be reduced and code can be simplified using the fact that the graph is bipartite and capacity of every edge is either 0 or 1. The idea is to use DFS traversal to find a job for an applicant (similar to augmenting path in Ford-Fulkerson). We call bpm() for every applicant, bpm() is the DFS based function that tries all possibilities to assign a job to the applicant.

In bpm(), we one by one try all jobs that an applicant ‘u’ is interested in until we find a job, or all jobs are tried without luck. For every job we try, we do following.

If a job is not assigned to anybody, we simply assign it to the applicant and return true. If a job is assigned to somebody else say x, then we recursively check whether x can be assigned some other job. To make sure that x doesn’t get the same job again, we mark the job ‘v’ as seen before we make recursive call for x. If x can get other job, we change the applicant for job ‘v’ and return true. We use an array maxR[0..N-1] that stores the applicants assigned to different jobs.

If bmp() returns true, then it means that there is an augmenting path in flow network and 1 unit of flow is added to the result in maxBPM().

## C++

`// A C++ program to find maximal ` `// Bipartite matching. ` `#include <iostream> ` `#include <string.h> ` `using` `namespace` `std; ` ` ` `// M is number of applicants ` `// and N is number of jobs ` `#define M 6 ` `#define N 6 ` ` ` `// A DFS based recursive function ` `// that returns true if a matching ` `// for vertex u is possible ` `bool` `bpm(` `bool` `bpGraph[M][N], ` `int` `u, ` ` ` `bool` `seen[], ` `int` `matchR[]) ` `{ ` ` ` `// Try every job one by one ` ` ` `for` `(` `int` `v = 0; v < N; v++) ` ` ` `{ ` ` ` `// If applicant u is interested in ` ` ` `// job v and v is not visited ` ` ` `if` `(bpGraph[u][v] && !seen[v]) ` ` ` `{ ` ` ` `// Mark v as visited ` ` ` `seen[v] = ` `true` `; ` ` ` ` ` `// If job 'v' is not assigned to an ` ` ` `// applicant OR previously assigned ` ` ` `// applicant for job v (which is matchR[v]) ` ` ` `// has an alternate job available. ` ` ` `// Since v is marked as visited in ` ` ` `// the above line, matchR[v] in the following ` ` ` `// recursive call will not get job 'v' again ` ` ` `if` `(matchR[v] < 0 || bpm(bpGraph, matchR[v], ` ` ` `seen, matchR)) ` ` ` `{ ` ` ` `matchR[v] = u; ` ` ` `return` `true` `; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` `return` `false` `; ` `} ` ` ` `// Returns maximum number ` `// of matching from M to N ` `int` `maxBPM(` `bool` `bpGraph[M][N]) ` `{ ` ` ` `// An array to keep track of the ` ` ` `// applicants assigned to jobs. ` ` ` `// The value of matchR[i] is the ` ` ` `// applicant number assigned to job i, ` ` ` `// the value -1 indicates nobody is ` ` ` `// assigned. ` ` ` `int` `matchR[N]; ` ` ` ` ` `// Initially all jobs are available ` ` ` `memset` `(matchR, -1, ` `sizeof` `(matchR)); ` ` ` ` ` `// Count of jobs assigned to applicants ` ` ` `int` `result = 0; ` ` ` `for` `(` `int` `u = 0; u < M; u++) ` ` ` `{ ` ` ` `// Mark all jobs as not seen ` ` ` `// for next applicant. ` ` ` `bool` `seen[N]; ` ` ` `memset` `(seen, 0, ` `sizeof` `(seen)); ` ` ` ` ` `// Find if the applicant 'u' can get a job ` ` ` `if` `(bpm(bpGraph, u, seen, matchR)) ` ` ` `result++; ` ` ` `} ` ` ` `return` `result; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `// Let us create a bpGraph ` ` ` `// shown in the above example ` ` ` `bool` `bpGraph[M][N] = {{0, 1, 1, 0, 0, 0}, ` ` ` `{1, 0, 0, 1, 0, 0}, ` ` ` `{0, 0, 1, 0, 0, 0}, ` ` ` `{0, 0, 1, 1, 0, 0}, ` ` ` `{0, 0, 0, 0, 0, 0}, ` ` ` `{0, 0, 0, 0, 0, 1}}; ` ` ` ` ` `cout << ` `"Maximum number of applicants that can get job is "` ` ` `<< maxBPM(bpGraph); ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// A Java program to find maximal ` `// Bipartite matching. ` `import` `java.util.*; ` `import` `java.lang.*; ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// M is number of applicants ` ` ` `// and N is number of jobs ` ` ` `static` `final` `int` `M = ` `6` `; ` ` ` `static` `final` `int` `N = ` `6` `; ` ` ` ` ` `// A DFS based recursive function that ` ` ` `// returns true if a matching for ` ` ` `// vertex u is possible ` ` ` `boolean` `bpm(` `boolean` `bpGraph[][], ` `int` `u, ` ` ` `boolean` `seen[], ` `int` `matchR[]) ` ` ` `{ ` ` ` `// Try every job one by one ` ` ` `for` `(` `int` `v = ` `0` `; v < N; v++) ` ` ` `{ ` ` ` `// If applicant u is interested ` ` ` `// in job v and v is not visited ` ` ` `if` `(bpGraph[u][v] && !seen[v]) ` ` ` `{ ` ` ` ` ` `// Mark v as visited ` ` ` `seen[v] = ` `true` `; ` ` ` ` ` `// If job 'v' is not assigned to ` ` ` `// an applicant OR previously ` ` ` `// assigned applicant for job v (which ` ` ` `// is matchR[v]) has an alternate job available. ` ` ` `// Since v is marked as visited in the ` ` ` `// above line, matchR[v] in the following ` ` ` `// recursive call will not get job 'v' again ` ` ` `if` `(matchR[v] < ` `0` `|| bpm(bpGraph, matchR[v], ` ` ` `seen, matchR)) ` ` ` `{ ` ` ` `matchR[v] = u; ` ` ` `return` `true` `; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` `return` `false` `; ` ` ` `} ` ` ` ` ` `// Returns maximum number ` ` ` `// of matching from M to N ` ` ` `int` `maxBPM(` `boolean` `bpGraph[][]) ` ` ` `{ ` ` ` `// An array to keep track of the ` ` ` `// applicants assigned to jobs. ` ` ` `// The value of matchR[i] is the ` ` ` `// applicant number assigned to job i, ` ` ` `// the value -1 indicates nobody is assigned. ` ` ` `int` `matchR[] = ` `new` `int` `[N]; ` ` ` ` ` `// Initially all jobs are available ` ` ` `for` `(` `int` `i = ` `0` `; i < N; ++i) ` ` ` `matchR[i] = -` `1` `; ` ` ` ` ` `// Count of jobs assigned to applicants ` ` ` `int` `result = ` `0` `; ` ` ` `for` `(` `int` `u = ` `0` `; u < M; u++) ` ` ` `{ ` ` ` `// Mark all jobs as not seen ` ` ` `// for next applicant. ` ` ` `boolean` `seen[] =` `new` `boolean` `[N] ; ` ` ` `for` `(` `int` `i = ` `0` `; i < N; ++i) ` ` ` `seen[i] = ` `false` `; ` ` ` ` ` `// Find if the applicant 'u' can get a job ` ` ` `if` `(bpm(bpGraph, u, seen, matchR)) ` ` ` `result++; ` ` ` `} ` ` ` `return` `result; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main (String[] args) ` ` ` `throws` `java.lang.Exception ` ` ` `{ ` ` ` `// Let us create a bpGraph shown ` ` ` `// in the above example ` ` ` `boolean` `bpGraph[][] = ` `new` `boolean` `[][]{ ` ` ` `{` `false` `, ` `true` `, ` `true` `, ` ` ` `false` `, ` `false` `, ` `false` `}, ` ` ` `{` `true` `, ` `false` `, ` `false` `, ` ` ` `true` `, ` `false` `, ` `false` `}, ` ` ` `{` `false` `, ` `false` `, ` `true` `, ` ` ` `false` `, ` `false` `, ` `false` `}, ` ` ` `{` `false` `, ` `false` `, ` `true` `, ` ` ` `true` `, ` `false` `, ` `false` `}, ` ` ` `{` `false` `, ` `false` `, ` `false` `, ` ` ` `false` `, ` `false` `, ` `false` `}, ` ` ` `{` `false` `, ` `false` `, ` `false` `, ` ` ` `false` `, ` `false` `, ` `true` `}}; ` ` ` `GFG m = ` `new` `GFG(); ` ` ` `System.out.println( ` `"Maximum number of applicants that can"` `+ ` ` ` `" get job is "` `+m.maxBPM(bpGraph)); ` ` ` `} ` `} ` |

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## Python

`# Python program to find ` `# maximal Bipartite matching. ` ` ` `class` `GFG: ` ` ` `def` `__init__(` `self` `,graph): ` ` ` ` ` `# residual graph ` ` ` `self` `.graph ` `=` `graph ` ` ` `self` `.ppl ` `=` `len` `(graph) ` ` ` `self` `.jobs ` `=` `len` `(graph[` `0` `]) ` ` ` ` ` `# A DFS based recursive function ` ` ` `# that returns true if a matching ` ` ` `# for vertex u is possible ` ` ` `def` `bpm(` `self` `, u, matchR, seen): ` ` ` ` ` `# Try every job one by one ` ` ` `for` `v ` `in` `range` `(` `self` `.jobs): ` ` ` ` ` `# If applicant u is interested ` ` ` `# in job v and v is not seen ` ` ` `if` `self` `.graph[u][v] ` `and` `seen[v] ` `=` `=` `False` `: ` ` ` ` ` `# Mark v as visited ` ` ` `seen[v] ` `=` `True` ` ` ` ` `'''If job 'v' is not assigned to ` ` ` `an applicant OR previously assigned ` ` ` `applicant for job v (which is matchR[v]) ` ` ` `has an alternate job available. ` ` ` `Since v is marked as visited in the ` ` ` `above line, matchR[v] in the following ` ` ` `recursive call will not get job 'v' again'''` ` ` `if` `matchR[v] ` `=` `=` `-` `1` `or` `self` `.bpm(matchR[v], ` ` ` `matchR, seen): ` ` ` `matchR[v] ` `=` `u ` ` ` `return` `True` ` ` `return` `False` ` ` ` ` `# Returns maximum number of matching ` ` ` `def` `maxBPM(` `self` `): ` ` ` `'''An array to keep track of the ` ` ` `applicants assigned to jobs. ` ` ` `The value of matchR[i] is the ` ` ` `applicant number assigned to job i, ` ` ` `the value -1 indicates nobody is assigned.'''` ` ` `matchR ` `=` `[` `-` `1` `] ` `*` `self` `.jobs ` ` ` ` ` `# Count of jobs assigned to applicants ` ` ` `result ` `=` `0` ` ` `for` `i ` `in` `range` `(` `self` `.ppl): ` ` ` ` ` `# Mark all jobs as not seen for next applicant. ` ` ` `seen ` `=` `[` `False` `] ` `*` `self` `.jobs ` ` ` ` ` `# Find if the applicant 'u' can get a job ` ` ` `if` `self` `.bpm(i, matchR, seen): ` ` ` `result ` `+` `=` `1` ` ` `return` `result ` ` ` ` ` `bpGraph ` `=` `[[` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `, ` `0` `], ` ` ` `[` `1` `, ` `0` `, ` `0` `, ` `1` `, ` `0` `, ` `0` `], ` ` ` `[` `0` `, ` `0` `, ` `1` `, ` `0` `, ` `0` `, ` `0` `], ` ` ` `[` `0` `, ` `0` `, ` `1` `, ` `1` `, ` `0` `, ` `0` `], ` ` ` `[` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `], ` ` ` `[` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `0` `, ` `1` `]] ` ` ` `g ` `=` `GFG(bpGraph) ` ` ` `print` `(` `"Maximum number of applicants that can get job is %d "` `%` `g.maxBPM()) ` ` ` `# This code is contributed by Neelam Yadav ` |

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## C#

`// A C# program to find maximal ` `// Bipartite matching. ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// M is number of applicants ` ` ` `// and N is number of jobs ` ` ` `static` `int` `M = 6; ` ` ` `static` `int` `N = 6; ` ` ` ` ` `// A DFS based recursive function ` ` ` `// that returns true if a matching ` ` ` `// for vertex u is possible ` ` ` `bool` `bpm(` `bool` `[,]bpGraph, ` `int` `u, ` ` ` `bool` `[]seen, ` `int` `[]matchR) ` ` ` `{ ` ` ` `// Try every job one by one ` ` ` `for` `(` `int` `v = 0; v < N; v++) ` ` ` `{ ` ` ` `// If applicant u is interested ` ` ` `// in job v and v is not visited ` ` ` `if` `(bpGraph[u, v] && !seen[v]) ` ` ` `{ ` ` ` `// Mark v as visited ` ` ` `seen[v] = ` `true` `; ` ` ` ` ` `// If job 'v' is not assigned to ` ` ` `// an applicant OR previously assigned ` ` ` `// applicant for job v (which is matchR[v]) ` ` ` `// has an alternate job available. ` ` ` `// Since v is marked as visited in the above ` ` ` `// line, matchR[v] in the following recursive ` ` ` `// call will not get job 'v' again ` ` ` `if` `(matchR[v] < 0 || bpm(bpGraph, matchR[v], ` ` ` `seen, matchR)) ` ` ` `{ ` ` ` `matchR[v] = u; ` ` ` `return` `true` `; ` ` ` `} ` ` ` `} ` ` ` `} ` ` ` `return` `false` `; ` ` ` `} ` ` ` ` ` `// Returns maximum number of ` ` ` `// matching from M to N ` ` ` `int` `maxBPM(` `bool` `[,]bpGraph) ` ` ` `{ ` ` ` `// An array to keep track of the ` ` ` `// applicants assigned to jobs. ` ` ` `// The value of matchR[i] is the ` ` ` `// applicant number assigned to job i, ` ` ` `// the value -1 indicates nobody is assigned. ` ` ` `int` `[]matchR = ` `new` `int` `[N]; ` ` ` ` ` `// Initially all jobs are available ` ` ` `for` `(` `int` `i = 0; i < N; ++i) ` ` ` `matchR[i] = -1; ` ` ` ` ` `// Count of jobs assigned to applicants ` ` ` `int` `result = 0; ` ` ` `for` `(` `int` `u = 0; u < M; u++) ` ` ` `{ ` ` ` `// Mark all jobs as not ` ` ` `// seen for next applicant. ` ` ` `bool` `[]seen = ` `new` `bool` `[N] ; ` ` ` `for` `(` `int` `i = 0; i < N; ++i) ` ` ` `seen[i] = ` `false` `; ` ` ` ` ` `// Find if the applicant ` ` ` `// 'u' can get a job ` ` ` `if` `(bpm(bpGraph, u, seen, matchR)) ` ` ` `result++; ` ` ` `} ` ` ` `return` `result; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main () ` ` ` `{ ` ` ` `// Let us create a bpGraph shown ` ` ` `// in the above example ` ` ` `bool` `[,]bpGraph = ` `new` `bool` `[,] ` ` ` `{{` `false` `, ` `true` `, ` `true` `, ` ` ` `false` `, ` `false` `, ` `false` `}, ` ` ` `{` `true` `, ` `false` `, ` `false` `, ` ` ` `true` `, ` `false` `, ` `false` `}, ` ` ` `{` `false` `, ` `false` `, ` `true` `, ` ` ` `false` `, ` `false` `, ` `false` `}, ` ` ` `{` `false` `, ` `false` `, ` `true` `, ` ` ` `true` `, ` `false` `, ` `false` `}, ` ` ` `{` `false` `, ` `false` `, ` `false` `, ` ` ` `false` `, ` `false` `, ` `false` `}, ` ` ` `{` `false` `, ` `false` `, ` `false` `, ` ` ` `false` `, ` `false` `, ` `true` `}}; ` ` ` `GFG m = ` `new` `GFG(); ` ` ` `Console.Write( ` `"Maximum number of applicants that can"` `+ ` ` ` `" get job is "` `+m.maxBPM(bpGraph)); ` ` ` `} ` `} ` ` ` `//This code is contributed by nitin mittal. ` |

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**Output :**

Maximum number of applicants that can get job is 5

You may like to see below also:

Hopcroft–Karp Algorithm for Maximum Matching | Set 1 (Introduction)

Hopcroft–Karp Algorithm for Maximum Matching | Set 2 (Implementation)

**References:**

http://www.cs.cornell.edu/~wdtseng/icpc/notes/graph_part5.pdf

http://www.youtube.com/watch?v=NlQqmEXuiC8

http://en.wikipedia.org/wiki/Maximum_matching

http://www.stanford.edu/class/cs97si/08-network-flow-problems.pdf

http://www.cs.princeton.edu/courses/archive/spring13/cos423/lectures/07NetworkFlowII-2×2.pdf

http://www.ise.ncsu.edu/fangroup/or766.dir/or766_ch7.pdf

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