greedy

Car Fueling Problem by Greedy Alogorithm (getting list index out of range)

孤人 提交于 2021-01-27 14:23:48
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

Is Kadane's Algorithm Greedy or Optimised DP?

耗尽温柔 提交于 2021-01-04 03:34:03
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

Is Kadane's Algorithm Greedy or Optimised DP?

徘徊边缘 提交于 2021-01-04 03:28:18
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

Is Kadane's Algorithm Greedy or Optimised DP?

淺唱寂寞╮ 提交于 2021-01-04 03:27:07
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

Is Kadane's Algorithm Greedy or Optimised DP?

夙愿已清 提交于 2021-01-04 03:25:46
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