greedy

问题 I have a small problem solving the Car fueling problem using the Greedy Algorithm. Problem Introduction You are going to travel to another city that is located 𝑑 miles away from your home city. Your car can travel at most 𝑚 miles on a full tank and you start with a full tank. Along your way, there are gas stations at distances stop1 stop2 . . . ,stopN from your home city. What is the minimum number of refills needed? Input: 950 400 4 200 375 550 750 Output: 2 What I've tried as of now def car

问题 I feel like Kadane's algorithm is a modified version of the true dynamic programming solution of maximum subarray problem.Why do I feel so? I feel because the way to calculate the maximum subarray can be taken by: for(i=0;i<N;i++) { DP[i][A[i]]=true; for(j= -ve maximum ;j<= +ve maximum ;j++) if(DP[i-1][j]) DP[i][j+A[i]]=true; } The recurrence being if it is possible to form j with a subarray ending at i-1 elements i can form j+A[i] using the i th element and also form A[i] alone by starting a

问题 I feel like Kadane's algorithm is a modified version of the true dynamic programming solution of maximum subarray problem.Why do I feel so? I feel because the way to calculate the maximum subarray can be taken by: for(i=0;i<N;i++) { DP[i][A[i]]=true; for(j= -ve maximum ;j<= +ve maximum ;j++) if(DP[i-1][j]) DP[i][j+A[i]]=true; } The recurrence being if it is possible to form j with a subarray ending at i-1 elements i can form j+A[i] using the i th element and also form A[i] alone by starting a

问题 I feel like Kadane's algorithm is a modified version of the true dynamic programming solution of maximum subarray problem.Why do I feel so? I feel because the way to calculate the maximum subarray can be taken by: for(i=0;i<N;i++) { DP[i][A[i]]=true; for(j= -ve maximum ;j<= +ve maximum ;j++) if(DP[i-1][j]) DP[i][j+A[i]]=true; } The recurrence being if it is possible to form j with a subarray ending at i-1 elements i can form j+A[i] using the i th element and also form A[i] alone by starting a

问题 I feel like Kadane's algorithm is a modified version of the true dynamic programming solution of maximum subarray problem.Why do I feel so? I feel because the way to calculate the maximum subarray can be taken by: for(i=0;i<N;i++) { DP[i][A[i]]=true; for(j= -ve maximum ;j<= +ve maximum ;j++) if(DP[i-1][j]) DP[i][j+A[i]]=true; } The recurrence being if it is possible to form j with a subarray ending at i-1 elements i can form j+A[i] using the i th element and also form A[i] alone by starting a

来源： https://stackoverflow.com/questions/62694219/minimum-number-of-platforms-required-for-a-railway-station

来源： https://stackoverflow.com/questions/62694219/minimum-number-of-platforms-required-for-a-railway-station

来源： https://stackoverflow.com/questions/62694219/minimum-number-of-platforms-required-for-a-railway-station

来源： https://stackoverflow.com/questions/19055840/finding-out-the-maximum-number-if-cost-is-associated-with-using-each-digit

来源： https://stackoverflow.com/questions/19055840/finding-out-the-maximum-number-if-cost-is-associated-with-using-each-digit