Question Name:LET’S BEGIN! SRM DSA ELAB

#include<iostream>
 
using namespace std;
 
#define MAX 1000005
 
int ans[MAX], count=0;
 
int solve(int X)
{
    //cout<<X<<endl;
    count++;
    if(ans[X]!=0){
        return ans[X];
    }
 
 
    int i, j, result=10*MAX, minResult = 10*MAX;
    if(X<2){
        return MAX*10;
    }
 
    result = solve(X-7) + 1;
    if(minResult > result){
        minResult = result;
    }
    result = solve(X-5) + 1;
    if(minResult > result){
        minResult = result;
    }
    result = solve(X-3) + 1;
    if(minResult > result){
        minResult = result;
    }
    result = solve(X-2) + 1;
    if(minResult > result){
        minResult = result;
    }
 
    ans[X] = minResult;
    //cout<<count<<" "<<X<<" = "<<ans[X]<<endl;
    return ans[X];
 
}
 
int main()
{
    int T;
    ans[2] = ans[3] = ans[5] = ans[7] = 1;
    cin>>T;
    while(T--)
    {
        int X, minstep=MAX*10;
        cin>>X;
        minstep = solve(X);
 
        if(minstep ==10*MAX )
            minstep=-1;
        cout<<minstep<<endl;
        //cout<<count<<endl;
 
    }
}

In-Depth Analysis of the Problem "LET’S BEGIN!"

This problem is essentially about determining the minimum number of single-digit prime numbers (2, 3, 5, 7) whose sum is equal to a given number $X$ . The problem requires finding the minimal way to achieve this for multiple test cases, given the constraints of $X$ being as large as $10^6$ .

We need to find an efficient solution that avoids excessive recursive depth or computation time due to the size of $X$ and the number of test cases.

Let's break the problem down in detail:

Understanding the Problem

The task is to find out the minimum number of single-digit primes that sum to a given number $X$ . The single-digit primes available are:

If it's impossible to represent $X$ as the sum of these primes, we should return -1.

Constraints:

$T$ (the number of test cases) can be as large as 100.
Each $X$ (the target sum) can be as large as $10^6$ .
We are asked to return the minimum number of single-digit primes whose sum equals $X$ .

Key Considerations:

Direct brute force approaches will likely fail due to the large constraint of $X$ , so dynamic programming or memoization must be employed to avoid redundant calculations.
Efficient recursion or iterative computation should be used to find the solution for each test case.

Solution Approach

1. Dynamic Programming with Memoization:

The key observation here is that we are trying to break down $X$ into smaller subproblems. At each step, we want to determine if subtracting one of the single-digit primes from $X$ can reduce it to zero using the fewest number of steps.

Thus, we can express the problem as:

solve(X) is the function that returns the minimum number of single-digit primes required to sum up to $X$ .

The solution can be thought of recursively:

If $X = 2, 3, 5, 7$ , the result is 1 (as these are the primes themselves).
For any $X$ , the minimum number of steps to reach $X$ is one more than the minimum number of steps to reach $X-2$ , $X-3$ , $X-5$ , or $X-7$ .

The base cases are:

$ans[2] = 1, ans[3] = 1, ans[5] = 1, ans[7] = 1$ , since they are primes.

For each new value of $X$ , the recursive relation is:

ans[X] = \min(ans[X-2], ans[X-3], ans[X-5], ans[X-7]) + 1

Where we check the smallest result among these possibilities.

If $X$ cannot be reached using the primes (i.e., if it goes below 2 in any recursive call), the function should return a large value, signaling that no valid combination of primes exists.

2. Recursive Function:

The recursive function is used to calculate the answer for a given $X$ while avoiding recalculations by storing the results in an array ans[]. If ans[X] is already computed, the function can return it directly to save time.

3. Base Cases and Edge Handling:

For any $X < 2$ , return an arbitrarily large number to indicate that no valid sum can be formed.
For cases where $X$ is already a prime (i.e., 2, 3, 5, 7), we return 1.

4. Optimization Considerations:

Memoization ensures that previously computed results are reused, leading to significant reductions in computation time for large $X$ .
Precomputing the results for all values of $X$ from 1 to $10^6$ could potentially optimize the solution further by eliminating recursion during test case evaluation.

Code Explanation


#include<iostream>
using namespace std;

#define MAX 1000005
int ans[MAX], count=0;

int solve(int X) {
    count++;
    if(ans[X] != 0) {
        return ans[X];  // Memoization: Return previously computed result
    }

    int result = 10 * MAX, minResult = 10 * MAX;
    if(X < 2) {
        return MAX * 10;  // Impossible to form X if less than 2
    }

    // Recursively check all possibilities by subtracting 7, 5, 3, and 2
    result = solve(X - 7) + 1;
    if(minResult > result) minResult = result;
    result = solve(X - 5) + 1;
    if(minResult > result) minResult = result;
    result = solve(X - 3) + 1;
    if(minResult > result) minResult = result;
    result = solve(X - 2) + 1;
    if(minResult > result) minResult = result;

    ans[X] = minResult;  // Store result for future use
    return ans[X];
}

int main() {
    int T;
    ans[2] = ans[3] = ans[5] = ans[7] = 1;  // Base cases: 2, 3, 5, 7 can be formed using 1 prime
    cin >> T;
    while(T--) {
        int X, minstep = MAX * 10;
        cin >> X;
        minstep = solve(X);

        if(minstep == 10 * MAX)
            minstep = -1;  // If no solution, return -1
        cout << minstep << endl;
    }
}

Detailed Breakdown of Code Logic

Global Array ans[MAX]:
- This array stores the minimum number of primes required to form a particular number $X$ . The array is pre-initialized to zero for all values.
- For base cases $X = 2, 3, 5, 7$ , the answer is set to 1.
Recursive Function solve(X):
- The function attempts to solve for $X$ by recursively subtracting one of the four single-digit primes (2, 3, 5, 7).
- If a result for $X$ is already computed (ans[X] != 0), it returns the stored result, saving computation time.
- For each recursive call, the function returns the minimum result found by subtracting each prime and adds 1 to the result (representing the step taken).
- If $X$ becomes less than 2, it returns a large number to signify that no valid sum can be formed.
Main Function:
- Reads $T$ , the number of test cases.
- For each test case, it reads $X$ and calls solve(X) to compute the minimum number of primes required to sum to $X$ .
- If no solution exists (i.e., the result is an impossibly large number), it returns -1.

Time and Space Complexity

Time Complexity:

The time complexity of this approach is approximately $O(X)$ per test case due to memoization. Once a result for a particular $X$ is computed, retrieving it is $O(1)$ . In the worst case scenario, this would be $O(10^6)$ .

Recursion depth is minimized thanks to memoization, and each number from 1 to $X$ is computed only once.

Space Complexity:

The space complexity is $O(X)$ , where $X$ is the maximum value for which we store results (i.e., $10^6$ ).

Edge Cases Considered

Small Values of $X$ : If $X$ is less than 2, it returns -1.
Impossible Cases: If no combination of primes can form $X$ , the function outputs -1.

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