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Sherlock and Cost

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  • zoyashah41306
     + 0 comments

    Locksmith Courses offer the perfect starting point for anyone looking to enter the security industry with confidence. These hands-on programs teach essential skills, from lock picking to advanced security systems. With rising interest in crime-solving, think Sherlock—but with real-world, practical knowledge. Cost varies by course level, but the investment pays off in job readiness and independence. Gain valuable skills and open doors to a secure and rewarding career path today.

  • nuemon20
     + 1 comment

    testcase=int(input()) for i in range(testcase): n=int(input()) b=[int(x) for x in input().split(' ')]
    maxi, max1 = 0, 0 for i in range(1,len(b)): curr, prev = b[i], b[i-1] newmaxi = max(maxi + abs(curr - prev), max1 + (curr - 1)) newmax1 = max(maxi + abs(1 - prev), max1) maxi, max1 = newmaxi, newmax1
    print(max(maxi, max1))

    • nuemon20
       + 0 comments

      python 3

  • wifivop476
     + 1 comment

    Sherlock and Cost, a leading provider of investigative services, offers unparalleled expertise in solving complex cases. With a team of seasoned professionals and cutting-edge technology, we deliver comprehensive solutions tailored to your needs Gatwick Airport Transfer. From corporate investigations to personal matters, trust Sherlock and Cost to uncover the truth. Plus, don't miss out on our exclusive Wow deals, providing exceptional value for our clients. Experience excellence in investigation services with Sherlock and Cost.

    • lengistonjorry
       + 0 comments

      Sherlock Holmes, the renowned fictional detective created by Sir Arthur Conan Doyle, was known for his exceptional deductive reasoning and keen observation skills, solving complex mysteries that often left others baffled. However, these brilliant investigative skills came at a cost. Holmes frequently engaged in perilous pursuits, risking his life to solve cases. His relentless focus on unearthing the truth often took a toll check here on his mental and physical well-being, and his friendship with Dr. Watson sometimes strained under the pressure.

  • mah_shamim
     + 2 comments

    Here's a PHP solution to solve this problem:

    <?php

function cost($B) {
    $n = count($B);

    // dp array where dp[0] means A[i] is 1 and dp[1] means A[i] is B[i]
    $dp = array_fill(0, 2, 0);

    for ($i = 1; $i < $n; $i++) {
        $prevDp0 = $dp[0];
        $prevDp1 = $dp[1];

        // When A[i] is 1
        $dp[0] = max(
            $prevDp0 + abs(1 - 1), // A[i-1] is 1
            $prevDp1 + abs(1 - $B[$i - 1]) // A[i-1] is B[i-1]
        );

        // When A[i] is B[i]
        $dp[1] = max(
            $prevDp0 + abs($B[$i] - 1), // A[i-1] is 1
            $prevDp1 + abs($B[$i] - $B[$i - 1]) // A[i-1] is B[i-1]
        );
    }

    // The result is the maximum value when A[n-1] is either 1 or B[n-1]
    return max($dp[0], $dp[1]);
}

// Example usage:
$B = [10, 1, 10, 1, 10];
echo cost($B); // Output: 36

?>

    Explanation:

    1. Dynamic Programming Table (dp):

      • dp[0]: The maximum cost when A[i] is set to 1.
      • dp[1]: The maximum cost when A[i] is set to B[i].

    2. Transition between states:

      • For each i from 1 to n-1, we update dp[0] and dp[1] based on the previous values:
        • If A[i] is 1, the cost can come from either A[i-1] being 1 or B[i-1].
        • If A[i] is B[i], the cost can come from either A[i-1] being 1 or B[i-1].

    3. Final Result:

      • The result will be the maximum value between dp[0] and dp[1] at the end of the array, which represents the maximum cost achievable by either setting A[n-1] to 1 or B[n-1].

    This solution efficiently computes the desired result with a time complexity of O(n), making it suitable for large inputs within the constraints.

    • cofagegaetaph
       + 0 comments

      Sherlock Holmes, the world-renowned detective, is often depicted as a figure whose work transcends monetary concerns. However, the cost of his services is sometimes referenced subtly throughout Sir Arthur Conan Doyle's stories. While Sherlock's cases often involve the wealthy, like in The Adventure of the Speckled Band, he is also known to take on cases out of sheer intellectual curiosity, regardless of the client's ability to pay. Dr. Watson mentions in some instances wakeboard bindings that Holmes waived his fee entirely for those in dire straits, revealing that his primary motivation was solving complex mysteries, rather than financial gain.

    • loius0044
       + 0 comments

      In Arthur Conan Doyle's tales, Sherlock Holmes often grappled with the complexities of cost, not just in monetary terms but also in terms of personal sacrifice and societal impact. Whether it was the financial ramifications of his various cases or the toll they took on his personal life, the concept of cost was a recurring theme in his adventures. For instance, Holmes's unwavering commitment to solving mysteries often led him to neglect his own well-being and relationships, underscoring the idea that true dedication KN Car Brand can come at a significant personal cost. Similarly, his cases frequently involved financial transactions, revealing the intricate ways in which crime and economics intersect. Through Holmes's experiences, Doyle explored how the pursuit of justice and the resolution of intricate problems involve weighing not just the immediate costs but also the broader, often intangible, repercussions.

  • marnalaashok8055
     + 4 comments
    #include<bits/stdc++.h>
#define fast_io ios_base::sync_with_stdio(false);cin.tie(NULL)
using namespace std;
typedef long long ll;
ll mod = 1e9+7;
ll dp[100001][101];
ll Fun(ll idx, ll n, vector<ll>&B,ll previous){
    if (idx == n)return 0;
    if(dp[idx][previous]!=-1)return dp[idx][previous];
    ll ans = 0 , left  = 0;
    if(idx == 0)ans = max(Fun(idx+1,n,B,1),Fun(idx+1,n,B,B[idx]));
    else{
        left = max(abs(previous-1) + Fun(idx+1,n,B,1), abs(previous-B[idx])+Fun(idx+1,n,B,B[idx]));
        ans = max(ans,left);
    }
    return dp[idx][previous]=ans;
}
void Don(){
    ll n;
    cin >> n;
    vector<ll>B(n);
    for(int i = 0 ; i < n ; i++)cin >> B[i];
    memset(dp,-1,sizeof(dp));
    ll ans = Fun(0,n,B,0);
    cout<<ans<<endl;
}
int main(){
    ll t;
    cin >> t;
    while(t--){
        Don();
    }
}
/*
instead of trying to place A[i] from 1 to B[i], just try 1 or B[i] because 1-B[i] gives us maximum.
here is my solution using recursion to try all possible ways of keeping either 1 or B[i]
after that i added memoization 
*/

    • 21P31A1216
       + 2 comments

      Thanks for this approach :) Very Understandable

    • jerrothomas790
       + 0 comments

      With Sherlock's sharp wit and deductive reasoning and Cost's culinary mastery and attention to detail, they delve into the intricacies of flavor, texture, and presentation through Culinary creations. From the bustling streets of Paris to the serene countryside of Tuscany, they traverse the globe in search of culinary enigmas waiting to be solved.

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