i am trying to solve a python algorithm problem but having trouble i will explain the question below ( I couldn't copy it here) Q : in a N*N array, each square consisting of random numbers, (every number is different) You are at a random number, you need to move to a neighboring number that is smaller than you. You can only move in 4 directions(up, down, left, right). When you move, if there are more than one place you can move to, you should only move to the smallest number possible. example: 19 57 74 73 94 26 27 32 98 61 40 88 49 38 25 21 66 53 95 46 80 23 58 39 89 if you start at 49, you move like this: 49 -> 32 -> 27 -> 26 -> 19, so you move 4 times. Keep moving until you cannot move anymore. Return the maximal times you can move in each test case. my code: import sys sys.stdin = open('input.txt', 'r') T = int(input()) for test_case in range(1, T + 1): N = int(input()) arr = [list(map(int, input().split())) for _ in range(N)] # directions di = [-1, 1, 0, 0] dj = [0, 0, -1, 1] # max move # max(count_list) count_list = [] count = 0 # count every time you move for i in range(N): for j in range(N): for d in range(4): ni = i + di[d] nj = j + dj[d] small_list_arr = [] # smallest of arr[ni][nj] if 0 <= ni < N and 0 <= nj < N: # inside N*N if arr[ni][nj] < arr[j]: # if there is smaller neighboring num small_list_arr.append(arr[ni][nj]) # if there are multiple smaller neighboring nums if arr[ni][nj] == min(small_list_arr): count += 1 # move count ni, i = i, ni # new i, j nj, j = j, nj count_list.append(count) print(f'#{test_case} {max(count_list)}') my problem is, I think I am okay with finding the new (i, j) (smallest neighboring number's location) but when I use the debugger I find that <for d in range(4)> does not start over from 0 again, for example if I found new (i, j) when d = 2, the debugger shows that when you go back up to <for d in range(4)> the d starts from 2. how do i solve this? maybe there is a problem with how i allocate ni, nj to i, j variables? I tried break but didn't work, I am new to python so need help sorry for my english Continue reading...
