<p>You have <code>n</code> dice, and each dice has <code>k</code> faces numbered from <code>1</code> to <code>k</code>.</p>

<p>Given three integers <code>n</code>, <code>k</code>, and <code>target</code>, return <em>the number of possible ways (out of the </em><code>k<sup>n</sup></code><em> total ways) </em><em>to roll the dice, so the sum of the face-up numbers equals </em><code>target</code>. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 1, k = 6, target = 3
<strong>Output:</strong> 1
<strong>Explanation:</strong> You throw one die with 6 faces.
There is only one way to get a sum of 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 2, k = 6, target = 7
<strong>Output:</strong> 6
<strong>Explanation:</strong> You throw two dice, each with 6 faces.
There are 6 ways to get a sum of 7: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 30, k = 30, target = 500
<strong>Output:</strong> 222616187
<strong>Explanation:</strong> The answer must be returned modulo 10<sup>9</sup> + 7.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n, k &lt;= 30</code></li>
	<li><code>1 &lt;= target &lt;= 1000</code></li>
</ul>

class Solution {
public:

    long long mod = 1000000007;
    long long  solveUsingMem(long long  n, long long  k, long long  target, vector<vector<long long >>&dp){
        if(target != 0 && n<=0 || (target<0)){
            return 0;
        }
        if(target == 0 && n == 0){
            return 1;
        }
        if(dp[n][target] != -1){
            return dp[n][target];
        }
        long long  count = 0;
        for(long long  i=1; i<=k; i++){
            long long temp = solveUsingMem(n-1, k, target-i,dp);
            count = (count%mod + temp%mod)%mod;
        }
        dp[n][target] = count;
        return dp[n][target];
    }
    long long  solveUsingTab(int n, int k, int target){
        vector<vector<long long >>dp(n+1,vector<long long >(target+1,0));           
        dp[0][0] = 1;
        for(int index=1; index<=n; index++){
            for(int j=1; j<=target; j++){
                long long  count = 0;
                for(long long  i=1; i<=k; i++){
                    if(j-i>=0){
                        long long temp = dp[index-1][j-i];
                        count = (count%mod + temp%mod)%mod;
                    }
                }
                dp[index][j] = count;
            }
        }
        return dp[n][target];
    }
    long long  solveUsingTabSO(int n, int k, int target){
        vector<long long>curr(target+1,0);
        vector<long long>prev(target+1,0);
        prev[0] = 1;
        for(int index=1; index<=n; index++){
            for(int j=1; j<=target; j++){
                long long  count = 0;
                for(long long  i=1; i<=k; i++){
                    if(j-i>=0){
                        long long temp = prev[j-i];
                        count = (count%mod + temp%mod)%mod;
                    }
                }
                curr[j] = count;
            }
            prev = curr;
        }
        return prev[target];
    }
    int numRollsToTarget(int n, int k, int target) {
        vector<vector<long long >>dp(n+1,vector<long long >(target+1,-1));
        long long temp = solveUsingTabSO(n,k,target);
        return temp%mod;
    }
};
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