Oct 2nd : Number of dice rolls with target sum

Author: vishwajeet-hogale

Dynamic Programming/number_of_dice_rolls_with_target_sum.cpp

@@ -0,0 +1,39 @@
+#include<iostream>
+#include<bits/stdc++.h>
+using namespace std;
+int dp[31][1001] = {0}; // Table of n * target sum ; This is used to save the results in bottom up approach
+int numRollsToTarget(int n, int k, int target) {
+    dp[0][0] = 1; // When you have 0 dices and target sum = 0; the number of ways to get the answer is 1
+    
+    for(int i = 1;i<=target;i++){ // when you have 0 dices; [1,targetsum] numbers there are zero ways to get the answer
+        dp[0][i] = 0;
+    }
+    for(int i=1;i<=n;i++){ // when you have 0 target; n =[1,n] dices there are zero ways to get the answer
+        dp[i][0] = 0;
+    }
+
+    /*
+    The below loop has Time complexity = O(n*targetSum*k)
+    It finds the answer for smaller target sums and build up to find the solution of the targetSum
+    */
+    for(int i=1;i<=n;i++){
+        for(int j=1;j<=target;j++){
+            long long res = 0;
+            for(int x=1;x<=k;x++){
+                if(j>=x){
+                    res = (res % 1000000007) + (dp[i-1][j-x] % 1000000007);
+                    // res = res % 1000000007;
+                }
+            }
+            dp[i][j] = res ;
+        }
+    }
+    return dp[n][target] % 1000000007;
+}
+
+int main(){
+
+    int n = 7,k=6,target = 30;
+    cout<<numRollsToTarget(n,k,target)<<endl;
+    return 0;
+}