Divide and conquer meaning
Recursively call the function by incrementing the start index and assign it to op2.Else, create two variables op1 and op2.If the characters at the start and end indices of the string are same, recursively provoke the function by incrementing the start index by 1 and decrementing the end index by 1.Else, if the startIndex is equal to endIndex, return 1.If the startIndex is greater than endIndex, return 0.Declare the function and take the strings s and the start and end indices as parameters.The Longest Palindromic Subsequence problem can be implemented using the Divide and Conquer Algorithm using the following steps: Implementation of Longest Palindromic Subsequence using Divide and Conquer Algorithm Print(findLCS("elephant", "eretpat", 0, 0)) #Function to solve Longest Common Subsequence Problem Return the maximum of the two variables op1 and op2.Recursively call the function by incrementing the index of s2 and assign it to op2.Recursively call the function by incrementing the index of s1 and assign it to op1.If the characters at the indices of the respective elements are same, recursively provoke the function by incrementing the indices by 1.If any of the indices of the strings is equal to its respective length, return 0.
The Longest Common Subsequence problem can be implemented using the Divide and Conquer Algorithm using the following steps: Implementation of Longest Common Subsequence using Divide and Conquer Algorithm Profit2 = zeroKnapsack(items, capacity, currentIndex+1) Profit1 = items.profit + zeroKnapsack(items, capacity-items.weight, currentIndex+1) #Function to solve Zero One Knapsack Problemĭef zeroKnapsack(items, capacity, currentIndex):Įlif items.weight <= capacity:
Return the maximum out of these two profits.This is achieved by recursively calling the function with different parameters. Else, if the total weight is less than the capacity, create two profits for checking adajacent items.If capacity is less than 0, currentIndex is less than 0, or currentIndex is greater than length of the list, return 0.Declare the function and take the items, capacity and current index as parameters.Zero One Knapsack Problem can be implemented using the Divide and Conquer Algorithm using the following steps: Implementation of Zero One Knapsack Problem using Divide and Conquer Algorithmįor the implementation of the Zero Knapsack Problem, we first create the Item where each object holds two variables – weight and profit. Print(findMinOperation("table", "tbres", 0, 0)) Return min (deleteOp, insertOp, replaceOp) ReplaceOp = 1 + findMinOperation(s1, s2, index1+1, index2+1) InsertOp = 1 + findMinOperation(s1, s2, index1+1, index2) Return findMinOperation(s1, s2, index1+1, index2+1)ĭeleteOp = 1 + findMinOperation(s1, s2, index1, index2+1) #Function to solve String Convert Problemĭef findMinOperation(s1, s2, index1, index2): Return the minimum of the three values as output.
Else, recursively invoke the function for each seperate operation by incrementing the value of the indices.If they have the same length, recursively invoke the function by incementing the index.If s1 or s2 are empty strings, return the differnece between the empty string and the other string as output.Declare the function and take the strings s1 and s2, along with their starting indices as parameters.String Convert problem can be implemented using the Divide and Conquer Algorithm using the following steps: Implementation of String Convert using Divide and Conquer Algorithm