Editing 2020 Jan Gold Problem 1 Time is Mooney (section)

== Solution ==

The solution to this problem can be solved using dynamic programming.

We create an array dp[N+1] where dp[i] represents the minimum amount of time needed to complete the first i tasks.

We can then fill the array using the following recurrence relation:

dp[i] = min(dp[i], dp[i-1] + time[i])

Where time[i] is the time required to complete the i-th task.

We can then iterate through the array and find the minimum amount of time needed to complete all tasks.

The following is a C++ implementation of the solution:

<pre>
int N; // number of tasks
int T; // total time available
int time[N+1]; // time required to complete each task
int dp[N+1]; // dp[i] represents the minimum amount of time needed to complete the first i tasks

dp[0] = 0;

for (int i = 1; i <= N; i++) {
    dp[i] = T + 1; // set dp[i] to a value greater than T
    for (int j = 0; j < i; j++) {
        dp[i] = min(dp[i], dp[j] + time[i]);
    }
}

int ans = T + 1;
for (int i = 0; i <= N; i++) {
    ans = min(ans, dp[i]);
}

if (ans > T) {
    cout << "Not possible" << endl;
} else {
    cout << ans << endl;
}
</pre>

[[Category:Yearly_2019_2020]]
[[Category:Gold]]
[[Category:Dynamic Programming]]
[[Category:Graph]]
[[Category:DFS]]
[[Category:Memoization]]