Solved DP-1 coinchange

CoinsChange.java

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+// Time Complexity:O(n*m)
+// Space complexity: O(n*m)
+//We build a 2D DP table where dp[i][j] is the min coins to make amount j using first i coins.
+// If we can't pick the current coin, we just use the value from the row above.
+// Otherwise, we take the min between skipping it and picking it (which adds 1 coin)
+
+class Solution {
+    public int coinChange(int[] coins, int amount) {
+        int rows = coins.length;
+        int columns = amount;
+        // declare a 2-D Array to capture the values of no of coins for a given amount
+        int[][] dp = new int [rows+1][columns+1];
+        // fill first row with value > total amount
+        for (int col = 0; col<= columns; col++) {
+            dp[0][col] = amount+1;
+        }
+        // fill the remaining rows and clolumns of the table
+        for(int i = 1; i <= rows; i++){
+
+            for(int j= 1; j<= columns; j++) {
+                if(j < coins[i-1]){
+                    dp[i][j] = dp[i-1][j];
+                } else {
+                    dp[i][j] = Math.min(dp[i-1][j], dp[i][j-coins[i-1]]+1);
+                }
+            }
+        }
+        if(dp[rows][columns] > amount) return -1;
+        return dp[rows][columns];
+    }
+}