diff --git a/README.md b/README.md
index b0e3967..c0d0852 100644
--- a/README.md
+++ b/README.md
@@ -27,6 +27,17 @@
| Solution |
Topics |
+
+ | September 7th |
+ 1155. Number of Dice Rolls With Target Sum |
+ $\text{\color{Dandelion}Medium}$ |
+
+ Dynamic Programming - Tabulation
+ |
+
+ Dynamic Programming
+ |
+
| September 6th |
1137. N-th Tribonacci Number |
@@ -77,18 +88,6 @@
Hash Table
-
- | September 2nd |
- 923. 3Sum With Multiplicity |
- $\text{\color{Dandelion}Medium}$ |
-
- Counting
- |
-
- Array,
- Counting
- |
-
@@ -4341,6 +4340,18 @@
|
+
+ | 1155 |
+ Number of Dice Rolls With Target Sum |
+ Java with Dynamic Programming using
+ Tabulation
+ |
+ $\text{\color{Dandelion}Medium}$ |
+
+ Dynamic Programming
+ |
+ |
+
| 1578 |
Minimum Time to Make Rope Colorful |
diff --git a/solutions/1155. Number of Dice Rolls With Target Sum/NumberOfDiceRollsWithTargetSum_Tabulation.java b/solutions/1155. Number of Dice Rolls With Target Sum/NumberOfDiceRollsWithTargetSum_Tabulation.java
new file mode 100644
index 0000000..ce0dfbe
--- /dev/null
+++ b/solutions/1155. Number of Dice Rolls With Target Sum/NumberOfDiceRollsWithTargetSum_Tabulation.java
@@ -0,0 +1,41 @@
+package com.cheehwatang.leetcode;
+
+// Time Complexity : O(n * k * target),
+// where 'n', 'k' and 'target' are the variables used by the method.
+// For each dice, 'n', we check each values of 'target' to add the number of faces 'k' that can sum to 'target',
+// in a 3 layer nested for-loops.
+//
+// Space Complexity : O(n * target),
+// where 'n' and 'target' are the variables used by the method.
+// We use a matrix of 'n' rows with 'target' columns for tabulation.
+
+public class NumberOfDiceRollsWithTargetSum_Tabulation {
+
+ // Approach:
+ // A dynamic programming problem where there are overlapping subproblem,
+ // for which each roll has possible 1 to k number face-up.
+ // Here, the iterative and tabulation method is used.
+ // By reducing the target by the face-up number each roll (hence n - 1),
+ // the successful sequence of rolls is achieved when n = 0 and target = 0, which forms the base case for the table.
+
+ public int numRollsToTarget(int n, int k, int target) {
+ // As the array is 0-indexed, we need + 1 length to accommodate 'n' and 'target' respectively.
+ int[][] table = new int[n + 1][target + 1];
+ // The seed/base case for successful sequence in the table.
+ table[0][0] = 1;
+ // Iterate through the table to get the result for all possibilities.
+ int modulus = (int) (1e9 + 7);
+ for (int i = 0; i < n; i++) {
+ for (int j = 0; j < target; j++) {
+ // Skip if the position in the array is 0, meaning nothing to add to the j + 1 to k positions.
+ if (table[i][j] == 0) continue;
+ for (int face = 1; face <= k; face++) {
+ if (face + j <= target) {
+ table[i + 1][j + face] = (table[i + 1][j + face] + table[i][j]) % modulus;
+ }
+ }
+ }
+ }
+ return table[n][target];
+ }
+}