Week2/day3

POD/week1-arrays/day4.py

@@ -0,0 +1,26 @@
+
+for _ in range(int(input())):    
+    n,s = map(int,input().split())
+    arr = list(map(int,input().split()))
+    #before start setting the indices of the subarray
+    i0,j0=-1,-1
+    for i in range(n):
+        #setting sum as the first element of the subarray
+        su = arr[i]
+        try:
+            #checking for an index j so that sum of arr[i:j] is the req sum
+            for j in range(i+1,n):
+                su += arr[j]
+                #if the sum is required sum set the indices of i and j
+                if su == s:
+                    i0 = i
+                    j0 = j
+                    break
+
+        except:
+            pass                    
+        if j0!= -1:
+            print(i0+1,j0+1)
+            break
+    else:
+        print(-1)

POD/week1-arrays/day5_2.py

@@ -0,0 +1,19 @@
+t = int(input())
+for _ in range(t):
+    n = int(input())
+    l = list(map(int,input().split()))
+    c = 0
+    #finding the bigger of the first and second element
+    p = max(l[:1])
+    for i in range(n):
+            #condition for record breaking:
+            #index 0: day 1 visitors is more than day 2
+            #day 2 to day n-1(day i): visitors is max of the subarray [0:i], and visitors on day i is more than i+1
+            #day n(last day) (index n-1):visitors is max of the full array 
+            if (i == 0 or l[i] > p) and (i == n-1 or l[i]> l[i+1]):               
+                c+=1
+            #updating the maximum no of visitors from the  subarray[0:i]
+            if l[i]>p:
+                p = l[i]
+
+    print("Case #{}: {}".format(_+1,c))

POD/week2/day3.py

@@ -0,0 +1,23 @@
+class Solution(object):
+    def minimumSwap(self, s1, s2):
+        x,y = 0,0
+        swaps = 0
+        for i in range(len(s1)):
+        #ignore all the x and y which are equal
+            if s1[i]!=s2[i]:
+                if s1[i]=="x":
+                    x+=1
+                else:
+                    y+=1               
+
+        #if the total number of x and y we need to swap is odd, we cant swap to form 2 equak strings
+        if ((x+y)%2 ==1):
+            return -1
+        else:
+            #we can group each to be swapped into pairs of xx or yy
+            #then it will be xx(s1) and yy(s2) or vice versa which requires 1 swap
+            #so for total xx(s1) and yy(s2) pairs we can get it as x//2 + x//y
+            #the remaining will be odd number of x(s1) and y(s2) or vice versa which requires 1 swap
+            #so for total xy and yx it is x%2(reminder when x//2) + y%2(reminder when y//2)
+
+            return y//2+y%2+x//2+x%2