710295. Minimize Cash Flow

Medium

Minimize Cash Flow

Summary

The Minimize Cash Flow problem involves minimizing a given cash flow by moving some of it to another account. The goal is to minimize the difference between two amounts, one in each account. This problem requires an understanding of dynamic programming and optimization techniques.

Detailed Explanation

We are given a series of transactions where we need to keep track of the amount of money in our accounts. For each transaction, we can either move some cash from one account to another or leave it as is. The goal is to minimize the difference between the amounts in both accounts.

The dynamic programming approach involves creating a table dp where dp[i][j] represents the minimum difference when we have processed the first i transactions and the current balance in each account is j. For each transaction, we calculate the new balances in both accounts and update our table accordingly.

Here's a step-by-step breakdown of the solution:

  1. Initialize an empty table dp with dimensions (transactions + 1) x (max_balance + 1) where transactions is the number of transactions and max_balance is the maximum balance in either account.
  2. Iterate through each transaction:
    • For each transaction, calculate the new balances in both accounts by adding/subtracting the transaction amount from the current balances.
    • Update our table dp with the minimum difference when we have processed this transaction and the new balances are j.
  3. Return the minimum difference found in the last row of the table.

Here's a sample diagram representing the transactions:

  +---------------+
  |  Transaction  |
  +---------------+
  | 0 | 100 | 50 |
  +---------------+
  | 1 | 150 | 75 |
  +---------------+
  | 2 | 180 | 90 |
  +---------------+
  ...

Time complexity: O(transactions * max_balance) Space complexity: O(transactions * max_balance)

Optimized Solutions

Java

View on GeeksforGeeks
public int minimizeCashFlow(int[] transactions) {
    int minDiff = Integer.MAX_VALUE;
    for (int i = 0; i < transactions.length - 1; i++) {
        int newBalance1 = 100 + transactions[i];
        int newBalance2 = 50 - transactions[i];
        minDiff = Math.min(minDiff, Math.abs(newBalance1 - newBalance2));
    }
    return minDiff;
}

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