710295. Minimize Cash Flow
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:
- Initialize an empty table
dpwith dimensions(transactions + 1) x (max_balance + 1)wheretransactionsis the number of transactions andmax_balanceis the maximum balance in either account. - 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
dpwith the minimum difference when we have processed this transaction and the new balances arej.
- 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
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;
}