Minimize Cash Flow among a given set of friends who have borrowed money from each other

Given a number of friends who have to give or take some amount of money from one another. Design an algorithm by which the total cash flow among all the friends is minimized.

Example:
Following diagram shows input debts to be settled.

Above debts can be settled in following optimized way

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

The idea is to use Greedy algorithm where at every step, settle all amounts of one person and recur for remaining n-1 persons.
How to pick the first person? To pick the first person, calculate the net amount for every person where net amount is obtained by subtracting all debts (amounts to pay) from all credits (amounts to be paid). Once net amount for every person is evaluated, find two persons with maximum and minimum net amounts. These two persons are the most creditors and debtors. The person with minimum of two is our first person to be settled and removed from list. Let the minimum of two amounts be x. We pay ‘x’ amount from the maximum debtor to maximum creditor and settle one person. If x is equal to the maximum debit, then maximum debtor is settled, else maximum creditor is settled.

The following is detailed algorithm.

Do following for every person Pi where i is from 0 to n-1.
1) Compute the net amount for every person. The net amount for person ‘i’ can be computed be subtracting sum of all debts from sum of all credits.

2) Find the two persons that are maximum creditor and maximum debtor. Let the maximum amount to be credited maximum creditor be maxCredit and maximum amount to be debited from maximum debtor be maxDebit. Let the maximum debtor be Pd and maximum creditor be Pc.

3) Find the minimum of maxDebit and maxCredit. Let minimum of two be x. Debit ‘x’ from Pd and credit this amount to Pc

4) If x is equal to maxCredit, then remove Pc from set of persons and recur for remaining (n-1) persons.

5)
If x is equal to maxDebit, then remove Pd from set of persons and recur for remaining (n-1) persons.

Thanks to Balaji S for suggesting this method in a comment here.

The following is implementation of above algorithm.

C++

 // C++ program to fin maximum cash flow among a set of persons #include using namespace std;    // Number of persons (or vertices in the graph) #define N 3    // A utility function that returns index of minimum value in arr[] int getMin(int arr[]) {     int minInd = 0;     for (int i=1; i arr[maxInd])             maxInd = i;     return maxInd; }    // A utility function to return minimum of 2 values int minOf2(int x, int y) {     return (x

Java

 // Java program to fin maximum cash  // flow among a set of persons    class GFG {     // Number of persons (or vertices in the graph)     static final int N = 3;            // A utility function that returns      // index of minimum value in arr[]     static int getMin(int arr[])     {         int minInd = 0;         for (int i = 1; i < N; i++)             if (arr[i] < arr[minInd])                 minInd = i;         return minInd;     }            // A utility function that returns      // index of maximum value in arr[]     static int getMax(int arr[])     {         int maxInd = 0;         for (int i = 1; i < N; i++)             if (arr[i] > arr[maxInd])                 maxInd = i;         return maxInd;     }            // A utility function to return minimum of 2 values     static int minOf2(int x, int y)     {         return (x < y) ? x: y;     }            // amount[p] indicates the net amount      // to be credited/debited to/from person 'p'     // If amount[p] is positive, then      // i'th person will amount[i]     // If amount[p] is negative, then      // i'th person will give -amount[i]     static void minCashFlowRec(int amount[])     {         // Find the indexes of minimum and         // maximum values in amount[]         // amount[mxCredit] indicates the maximum amount          // to be given (or credited) to any person .         // And amount[mxDebit] indicates the maximum amount          // to be taken(or debited) from any person.         // So if there is a positive value in amount[],          // then there must be a negative value         int mxCredit = getMax(amount), mxDebit = getMin(amount);                // If both amounts are 0, then          // all amounts are settled         if (amount[mxCredit] == 0 && amount[mxDebit] == 0)             return;                // Find the minimum of two amounts         int min = minOf2(-amount[mxDebit], amount[mxCredit]);         amount[mxCredit] -= min;         amount[mxDebit] += min;                // If minimum is the maximum amount to be         System.out.println("Person " + mxDebit + " pays " + min                                 + " to " + "Person " + mxCredit);                // Recur for the amount array.          // Note that it is guaranteed that         // the recursion would terminate          // as either amount[mxCredit]  or          // amount[mxDebit] becomes 0         minCashFlowRec(amount);     }            // Given a set of persons as graph[]      // where graph[i][j] indicates     // the amount that person i needs to      // pay person j, this function     // finds and prints the minimum      // cash flow to settle all debts.     static void minCashFlow(int graph[][])     {         // Create an array amount[],          // initialize all value in it as 0.         int amount[]=new int[N];                // Calculate the net amount to          // be paid to person 'p', and         // stores it in amount[p]. The          // value of amount[p] can be         // calculated by subtracting          // debts of 'p' from credits of 'p'         for (int p = 0; p < N; p++)         for (int i = 0; i < N; i++)             amount[p] += (graph[i][p] - graph[p][i]);                minCashFlowRec(amount);     }            // Driver code     public static void main (String[] args)     {         // graph[i][j] indicates the amount          // that person i needs to pay person j         int graph[][] = { {0, 1000, 2000},                             {0, 0, 5000},                             {0, 0, 0},};                // Print the solution         minCashFlow(graph);     } }    // This code is contributed by Anant Agarwal.

Python3

 # Python3 program to fin maximum # cash flow among a set of persons    # Number of persons(or vertices in graph) N = 3    # A utility function that returns # index of minimum value in arr[] def getMin(arr):            minInd = 0     for i in range(1, N):         if (arr[i] < arr[minInd]):             minInd = i     return minInd    # A utility function that returns # index of maximum value in arr[] def getMax(arr):        maxInd = 0     for i in range(1, N):         if (arr[i] > arr[maxInd]):             maxInd = i     return maxInd    # A utility function to # return minimum of 2 values def minOf2(x, y):        return x if x < y else y    # amount[p] indicates the net amount to # be credited/debited to/from person 'p' # If amount[p] is positive, then i'th  # person will amount[i] # If amount[p] is negative, then i'th # person will give -amount[i] def minCashFlowRec(amount):        # Find the indexes of minimum     # and maximum values in amount[]     # amount[mxCredit] indicates the maximum     # amount to be given(or credited) to any person.     # And amount[mxDebit] indicates the maximum amount     # to be taken (or debited) from any person.     # So if there is a positive value in amount[],      # then there must be a negative value     mxCredit = getMax(amount)     mxDebit = getMin(amount)        # If both amounts are 0,      # then all amounts are settled     if (amount[mxCredit] == 0 and amount[mxDebit] == 0):         return 0        # Find the minimum of two amounts     min = minOf2(-amount[mxDebit], amount[mxCredit])     amount[mxCredit] -=min     amount[mxDebit] += min        # If minimum is the maximum amount to be     print("Person " , mxDebit , " pays " , min         , " to " , "Person " , mxCredit)        # Recur for the amount array. Note that     # it is guaranteed that the recursion     # would terminate as either amount[mxCredit]      # or amount[mxDebit] becomes 0     minCashFlowRec(amount)    # Given a set of persons as graph[] where # graph[i][j] indicates the amount that # person i needs to pay person j, this # function finds and prints the minimum  # cash flow to settle all debts. def minCashFlow(graph):        # Create an array amount[],     # initialize all value in it as 0.     amount = [0 for i in range(N)]        # Calculate the net amount to be paid     # to person 'p', and stores it in amount[p].     # The value of amount[p] can be calculated by     # subtracting debts of 'p' from credits of 'p'     for p in range(N):         for i in range(N):             amount[p] += (graph[i][p] - graph[p][i])        minCashFlowRec(amount)        # Driver code    # graph[i][j] indicates the amount # that person i needs to pay person j graph = [ [0, 1000, 2000],           [0, 0, 5000],           [0, 0, 0] ]    minCashFlow(graph)    # This code is contributed by Anant Agarwal.

C#

 // C# program to fin maximum cash  // flow among a set of persons using System;    class GFG {     // Number of persons (or      // vertices in the graph)     static int N = 3;            // A utility function that returns      // index of minimum value in arr[]     static int getMin(int []arr)     {         int minInd = 0;         for (int i = 1; i < N; i++)             if (arr[i] < arr[minInd])                 minInd = i;         return minInd;     }            // A utility function that returns      // index of maximum value in arr[]     static int getMax(int []arr)     {         int maxInd = 0;         for (int i = 1; i < N; i++)             if (arr[i] > arr[maxInd])                 maxInd = i;         return maxInd;     }            // A utility function to return     // minimum of 2 values     static int minOf2(int x, int y)     {         return (x < y) ? x: y;     }            // amount[p] indicates the net amount      // to be credited/debited to/from person 'p'     // If amount[p] is positive, then      // i'th person will amount[i]     // If amount[p] is negative, then      // i'th person will give -amount[i]     static void minCashFlowRec(int []amount)     {         // Find the indexes of minimum and         // maximum values in amount[]         // amount[mxCredit] indicates the maximum amount          // to be given (or credited) to any person .         // And amount[mxDebit] indicates the maximum amount          // to be taken(or debited) from any person.         // So if there is a positive value in amount[],          // then there must be a negative value         int mxCredit = getMax(amount), mxDebit = getMin(amount);                // If both amounts are 0, then          // all amounts are settled         if (amount[mxCredit] == 0 &&              amount[mxDebit] == 0)             return;                // Find the minimum of two amounts         int min = minOf2(-amount[mxDebit], amount[mxCredit]);         amount[mxCredit] -= min;         amount[mxDebit] += min;                // If minimum is the maximum amount to be         Console.WriteLine("Person " + mxDebit +                            " pays " + min + " to " +                            "Person " + mxCredit);                // Recur for the amount array.          // Note that it is guaranteed that         // the recursion would terminate          // as either amount[mxCredit] or          // amount[mxDebit] becomes 0         minCashFlowRec(amount);     }            // Given a set of persons as graph[]      // where graph[i][j] indicates     // the amount that person i needs to      // pay person j, this function     // finds and prints the minimum      // cash flow to settle all debts.     static void minCashFlow(int [,]graph)     {         // Create an array amount[],          // initialize all value in it as 0.         int []amount=new int[N];                // Calculate the net amount to          // be paid to person 'p', and         // stores it in amount[p]. The          // value of amount[p] can be         // calculated by subtracting          // debts of 'p' from credits of 'p'         for (int p = 0; p < N; p++)         for (int i = 0; i < N; i++)             amount[p] += (graph[i,p] - graph[p,i]);                minCashFlowRec(amount);     }            // Driver code     public static void Main ()     {         // graph[i][j] indicates the amount          // that person i needs to pay person j         int [,]graph = { {0, 1000, 2000},                          {0, 0, 5000},                          {0, 0, 0},};                // Print the solution         minCashFlow(graph);     } }    // This code is contributed by nitin mittal.

Output:

Person 1 pays 4000 to Person 2
Person 0 pays 3000 to Person 2

Time Complexity: O(N2) where N is the number of persons.

tags:

Graph Greedy Greedy Graph