On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:


Input: [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.


Example 2:


Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation:
 0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.


Note:


       2 <= N <= 50.
       grid[i][j] is a permutation of [0, ..., N*N - 1].

Difficulty: MEDIUM_HARD
Tags: Graph traversal

Spanning trees

Sorting

Matrices

Data structures
Skill types: Sorting

Data structures

Starter code:
class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        

Solutions:
class Solution:

	def reorganizeString(self, S):
		counter = collections.Counter(S)
		h = [(-v, k) for (k, v) in list(counter.items())]
		heapq.heapify(h)
		ans = []
		while h:
			(v, c) = heapq.heappop(h)
			ans.append(c)
			counter[c] -= 1
			if len(ans) >= 2:
				ch = ans[-2]
				if counter[ch] > 0:
					heapq.heappush(h, (-counter[ch], ch))
		if len(ans) < len(S):
			return ''
		return ''.join(ans)


class Solution:

	def reorganizeString(self, S):
		from math import ceil
		dict1 = {}
		s = ''
		for i in range(len(S)):
			if S[i] not in dict1:
				dict1[S[i]] = 1
			else:
				dict1[S[i]] += 1
			if dict1[S[i]] > ceil(len(S) / 2):
				return ''
		dict2 = [item for item in list(dict1.items())]
		dict2.sort(key=lambda x: x[1], reverse=True)
		for i in range(dict2[0][1]):
			s = s + dict2[0][0]
		for i in range(1, len(dict2)):
			temp = 0
			value = dict2[i][1]
			key = dict2[i][0]
			while value > 0:
				for j in range(i):
					while value > 0 and dict2[j][0] + dict2[j][0] in s:
						s = s[0:s.index(dict2[j][0] + dict2[j][0])] + dict2[j][0] + key + dict2[j][0] + s[s.index(dict2[j][0] + dict2[j][0]) + 2:]
						value -= 1
				if value > 0:
					if key in s:
						temp = s.index(key)
						if temp >= 2:
							s = s[0:temp - 2] + key + s[temp - 2:]
						else:
							s = key + s
						value -= 1
					else:
						s = s + key
						value -= 1
		return s


class Solution:

	def reorganizeString(self, S):
		count = {}
		limit = (len(S) + 1) / 2
		for c in S:
			if c in count:
				count[c] += 1
			else:
				count[c] = 1
			if count[c] > limit:
				return ''
		pq = []
		for x in count.keys():
			c = count[x]
			pq.append((-c, x))
		heapq.heapify(pq)
		result = []
		while len(pq) >= 2:
			(count1, char1) = heapq.heappop(pq)
			(count2, char2) = heapq.heappop(pq)
			result.extend([char1, char2])
			if count1 != -1:
				heapq.heappush(pq, (count1 + 1, char1))
			if count2 != -1:
				heapq.heappush(pq, (count2 + 1, char2))
		answer = ''.join(result)
		if len(pq) == 1:
			answer += heapq.heappop(pq)[1]
		return answer


class Solution(object):

	def reorganizeString(self, S):
		N = len(S)
		A = []
		for (c, x) in sorted(((S.count(x), x) for x in set(S))):
			if c > (N + 1) / 2:
				return ''
			A.extend(c * x)
			print(A)
		ans = [None] * N
		(ans[::2], ans[1::2]) = (A[int(N / 2):], A[:int(N / 2)])
		print(ans[::2])
		return ''.join(ans)


class Solution:

	def reorganizeString(self, S):
		c = collections.Counter(S)
		if max(c.values()) <= (len(S) + 1) / 2:
			res = ''
			while c:
				out = c.most_common(2)
				if len(out):
					res += out[0][0]
					c[out[0][0]] -= 1
				if len(out) > 1:
					res += out[1][0]
					c[out[1][0]] -= 1
				c += collections.Counter()
			return res
		return ''


class Solution:

	def reorganizeString(self, S):
		cnt = collections.Counter(S)
		res = '#'
		while cnt:
			stop = True
			for (k, v) in cnt.most_common():
				if k != res[-1]:
					stop = False
					res += k
					cnt[k] -= 1
					if not cnt[k]:
						del cnt[k]
					break
			if stop == True:
				break
		return res[1:] if len(res) == len(S) + 1 else ''