Add BFS and CC Long Challenge Solutions

Author: cyberShaw

CodeChef/October Long Challenge/Chef-and-Easy-Queries.py

@@ -0,0 +1,19 @@
+for _ in range(int(input())):
+    n, k = map(int, input().split())
+    inp = list(map(int, input().split()))
+    if n == 1:
+        print(int((inp[0] / k) + 1))
+        continue
+    total = 0
+    day = 0
+    dayFound = 0
+    for i in range(n):
+        day += 1
+        if(inp[i] + total < k):
+            print(int(day))
+            dayFound = 1
+            break
+        total = (inp[i] + total) - k
+    if not dayFound:
+        day += total / k
+        print(int(day + 1))

CodeChef/October Long Challenge/Covid-Run.py

@@ -0,0 +1,15 @@
+for _ in range(int(input())):
+    N, K, X, Y = map(int, input().split())
+    z = X
+    f = 0
+    while True:
+        X = (X + K) % N
+        if Y == X:
+            f = 1
+            break
+        elif X == z:
+            break
+    if f == 1:
+        print("YES")
+    else:
+        print("NO")

CodeChef/October Long Challenge/Positive-AND.py

@@ -0,0 +1,31 @@
+def power(n):
+    return n != 0 and ((n & (n-1)) == 0) 
+
+def main():
+    for i in range(int(input())):
+        n = int(input())
+        if n == 1:
+            print(1)
+            continue
+        elif n == 3:
+            print("1 3 2")
+            continue
+        elif n == 5:
+            print("2 3 1 5 4")
+            continue
+        elif(power(n)):
+            print(-1)
+            continue
+        else:
+            print("2 3 1 5 4")
+            i = 6
+            while(i <= n):
+                if power(i):
+                    print(str(i+1) + " " + str(i), end=" ")
+                    i += 2
+                else:
+                    print(i)
+                    i += 1
+            print("\n")
+
+main()

DataStructures/Graph/bfs.py

@@ -0,0 +1,14 @@
+def bfs(g,N):
+    queue = []
+    visited = [False] * (N)
+    bfs = []
+    queue.append(0)
+    visited[0] = True
+    while queue:
+        cur = queue.pop(0)
+        bfs.append(cur)
+        for neighbour in g[cur]:
+            if not visited[neighbour]:
+                queue.append(neighbour)
+                visited[neighbour] = True
+    return bfs