Amazon Flipkart GreyOrange Microsoft Mobicip Morgan-Stanley Snapdeal Visa Payu

Author: rahulcode22

0-1Knapsack.py

@@ -0,0 +1,18 @@
+#Complexity O(nW)
+def knapsack(W, wt, val, n):
+    dp = [[0 for i in range(W+1)] for i in range(n+1)]
+
+    for i in range(n+1):
+        for j in range(W+1):
+            if i == 0 or j == 0:
+                dp[i][j] = 0
+            elif j < wt[i-1]:
+                dp[i][j] = dp[i-1][j]
+            else:
+                dp[i][j] = max(val[i-1] + dp[i-1][j-wt[i-1]], dp[i-1][j])
+    return dp[n][W]
+val = [60, 100, 120]
+wt = [10, 20, 30]
+W = 50
+n = len(val)
+print knapsack(W, wt, val, n)

interleavingString.py

@@ -22,3 +22,30 @@ def isInterleaved(a,b,c):
     print c + "is Interleaved of "+ a + " and " + b
 else:
     print "Not Interleaved"
+
+
+#Dynamic Approach
+def isInterleaved(a,b,c):
+    m = len(a)
+    n = len(b)
+
+    table = [[False for i in range(n+1)] for i in range(m+1)]
+
+    if m+n != len(c):
+        return False
+
+    for i in range(m+1):
+        for j in range(n+1):
+            if i == 0 and j == 0:
+                table[i][j] = True
+            elif i == 0 and (b[j-1] == c[j-1]):
+                table[i][j] = table[i][j-1]
+            elif j == 0 and a[i-1] == c[i-1]:
+                table[i][j] = table[i-1][j]
+            elif a[i-1] == c[i+j-1] and b[j-1] != c[i+j-1]:
+                table[i][j] = table[i-1][j]
+            elif a[i-1] != c[i+j-1] and b[j-1] == c[i+j-1]:
+                table[i][j] = table[i][j-1]
+            elif a[i-1] == c[i+j-1] and b[j-1] == c[i+j-1]:
+                table[i][j] = table[i-1][j] or table[i][j-1]
+    return table[m][n]

lengthOfLongestSubsequence.py

@@ -1,44 +1,19 @@
-def lbs(arr):
-    n = len(arr)
-    lis = [1 for i in range(n+1)]
+def lcs(s1, s2):
+    m = len(s1)
+    n = len(s2)
 
-    for i in range(1,n):
-        for j in range(0, i):
-            if (arr[i] > arr[j]) and lis[i] < lis[j] + 1:
-                lis[i] = lis[j]+1
-    lds = [1 for i in range(n+1)]
+    dp = [[None]*(n+1) for i in range(m+1)]
 
-    for i in reversed(range(n-1)):#loop from n-2 to 0
-        for j in reversed(range(i-1,n)):#loop from n-1 downto i-1
-            if arr[i] > arr[j] and lds[i] < lds[j]+1:
-                lds[i] = lds[j]+1
+    for i in range(m+1):
+        for j in range(n+1):
+            if i == 0 or j == 0:
+                dp[i][j] = 0
+            elif s1[i-1] == s2[j-1]:
+                dp[i][j] = dp[i-1][j-1] + 1
+            else:
+                dp[i][j] = max(dp[i-1][j],dp[i][j-1])
 
-    maximum = lis[0] + lds[0] - 1
-    for i in range(1, n):
-        maximum = max((lis[i]+lds[i]-1),maximum)
-    return maximum
-
-arr =  [ 64, 263, 11, 188, 396, 268, 116, 453, 207, 39, 312, 401, 396, 476, 56, 277, 104, 213, 79, 128, 36, 393, 117, 298, 371, 73, 110, -3, 350, 144, 210, 491, 312, 351, 183, 25, 230, 303, 478, 476, 276, 127, 184, 390, 481, 63, 270, 417, 257, 189, 18, 53, 80, 122, 358, 469, 244, 144, 152, 83, 188, 174, 381, 499, 280, 67, -1, 95, 489, 428, 302, 327, 25, 419, 207, 499, 457, 239, 12, 438, 54, 473, 116, 355, 428, -8, 37, 292, 116, 361, 186, 396, 375, -8, 416, 21, -8, 231, 51, 271, 318, 214, 361, 196, 118, 434, 133, 366, 192, 244, 400, 88, 444, 251, 165, 286, 15, 0, 223, 74, 84, 456, 464, 50, 209, 137, 67, 74, 4, 397, 403, 156, 70, 39, 187, 483, 267, 187, 464, 361, 74, 398, 479, 107, 123, 358, 317, 209, 192, 157, 61, 437, 330, -2, 276, 418, 418, 286, 303, 335, 317, 394, 193, 252, 64, 71, 322, 397, 494, 480, 285, 25, 298, 227, 16, 382, 452, 75, 304, 172, 469, 334, 294, 181, 394, 163, 326, 169, 30, 39, 425, 105, 294, 354, 207, 314, 386, 494 ]
-print "Length of LBS is",lbs(arr)
-
-#Another Approach
-class Solution:
-    # @param A : tuple of integers
-    # @return an integer
-    def longestSubsequenceLength(self, A):
-        m=0
-        L=[1]*len(A)
-        for i in range(1,len(A)):
-            for j in range(i):
-                if A[j]<A[i]:
-                    L[i]=max(L[i],L[j]+1)
-        R=[1]*len(A)
-        for i in range(len(A)-2,-1,-1):
-            for j in range(len(A)-1,i,-1):
-                if A[j]<A[i]:
-                    R[i]=max(R[i],R[j]+1)
-        for i in range(len(A)):
-            val=L[i]+R[i]-1
-            if val>m:
-                m=val
-        return m
+    return dp[m][n]
+x = "AGGTAB"
+y = "GXTXAYB"
+print "Length of LCS is ", lcs(x, y)