codeparrot/xlcost-text-to-code huggingface dataset:
Find subarray with given sum | Set 1 ( Nonnegative Numbers ) | An efficient program to print subarray with sum as given sum ; Returns true if the there is a subarray of arr [ ] with a sum equal to ' sum ' otherwise returns false . Also , prints the result ; Initialize curr_sum as value of first element and starting point as 0 ; Add elements one by one to curr_sum and if the curr_sum exceeds the sum , then remove starting element ; If curr_sum exceeds the sum , then remove the starting elements ; If curr_sum becomes equal to sum , then return true ; Add this element to curr_sum ; If we reach here , then no subarray ; Driver program to test above function
#include <stdio.h> NEW_LINE int subArraySum ( int arr [ ] , int n , int sum ) { int curr_sum = arr [ 0 ] , start = 0 , i ; for ( i = 1 ; i <= n ; i ++ ) { while ( curr_sum > sum && start < i - 1 ) { curr_sum = curr_sum - arr [ start ] ; start ++ ; } if ( curr_sum == sum ) { printf ( " Sum ▁ found ▁ between ▁ indexes ▁ % d ▁ and ▁ % d " , start , i - 1 ) ; return 1 ; } if ( i < n ) curr_sum = curr_sum + arr [ i ] ; } printf ( " No ▁ subarray ▁ found " ) ; return 0 ; } int main ( ) { int arr [ ] = { 15 , 2 , 4 , 8 , 9 , 5 , 10 , 23 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int sum = 23 ; subArraySum ( arr , n , sum ) ; return 0 ; }
Find a triplet that sum to a given value |  ; returns true if there is triplet with sum equal to ' sum ' present in A [ ] . Also , prints the triplet ; Fix the first element as A [ i ] ; Fix the second element as A [ j ] ; Now look for the third number ; If we reach here , then no triplet was found ; Driver program to test above function
#include <stdio.h> NEW_LINE bool find3Numbers ( int A [ ] , int arr_size , int sum ) { int l , r ; for ( int i = 0 ; i < arr_size - 2 ; i ++ ) { for ( int j = i + 1 ; j < arr_size - 1 ; j ++ ) { for ( int k = j + 1 ; k < arr_size ; k ++ ) { if ( A [ i ] + A [ j ] + A [ k ] == sum ) { printf ( " Triplet ▁ is ▁ % d , ▁ % d , ▁ % d " , A [ i ] , A [ j ] , A [ k ] ) ; return true ; } } } } return false ; } int main ( ) { int A [ ] = { 1 , 4 , 45 , 6 , 10 , 8 } ; int sum = 22 ; int arr_size = sizeof ( A ) / sizeof ( A [ 0 ] ) ; find3Numbers ( A , arr_size , sum ) ; return 0 ; }
Analysis of Algorithms | Set 2 ( Worst , Average and Best Cases ) | C implementation of the approach ; Linearly search x in arr [ ] . If x is present then return the index , otherwise return - 1 ; Driver program to test above functions
#include <stdio.h> NEW_LINE int search ( int arr [ ] , int n , int x ) { int i ; for ( i = 0 ; i < n ; i ++ ) { if ( arr [ i ] == x ) return i ; } return -1 ; } int main ( ) { int arr [ ] = { 1 , 10 , 30 , 15 } ; int x = 30 ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; printf ( " % d ▁ is ▁ present ▁ at ▁ index ▁ % d " , x , search ( arr , n , x ) ) ; getchar ( ) ; return 0 ; }
Binary Search | C program to implement recursive Binary Search ; A recursive binary search function . It returns location of x in given array arr [ l . . r ] is present , otherwise - 1 ; If the element is present at the middle itself ; If element is smaller than mid , then it can only be present in left subarray ; Else the element can only be present in right subarray ; We reach here when element is not present in array ; Driver method to test above
#include <stdio.h> NEW_LINE int binarySearch ( int arr [ ] , int l , int r , int x ) { if ( r >= l ) { int mid = l + ( r - l ) / 2 ; if ( arr [ mid ] == x ) return mid ; if ( arr [ mid ] > x ) return binarySearch ( arr , l , mid - 1 , x ) ; return binarySearch ( arr , mid + 1 , r , x ) ; } return -1 ; } int main ( void ) { int arr [ ] = { 2 , 3 , 4 , 10 , 40 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int x = 10 ; int result = binarySearch ( arr , 0 , n - 1 , x ) ; ( result == -1 ) ? printf ( " Element ▁ is ▁ not ▁ present ▁ in ▁ array " ) : printf ( " Element ▁ is ▁ present ▁ at ▁ index ▁ % d " , result ) ; return 0 ; }
Binary Search | C program to implement iterative Binary Search ; A iterative binary search function . It returns location of x in given array arr [ l . . r ] if present , otherwise - 1 ; Check if x is present at mid ; If x greater , ignore left half ; If x is smaller , ignore right half ; if we reach here , then element was not present ; Driver method to test above
#include <stdio.h> NEW_LINE int binarySearch ( int arr [ ] , int l , int r , int x ) { while ( l <= r ) { int m = l + ( r - l ) / 2 ; if ( arr [ m ] == x ) return m ; if ( arr [ m ] < x ) l = m + 1 ; else r = m - 1 ; } return -1 ; } int main ( void ) { int arr [ ] = { 2 , 3 , 4 , 10 , 40 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int x = 10 ; int result = binarySearch ( arr , 0 , n - 1 , x ) ; ( result == -1 ) ? printf ( " Element ▁ is ▁ not ▁ present " " ▁ in ▁ array " ) : printf ( " Element ▁ is ▁ present ▁ at ▁ " " index ▁ % d " , result ) ; return 0 ; }
Interpolation Search | C program to implement interpolation search with recursion ; If x is present in arr [ 0. . n - 1 ] , then returns index of it , else returns - 1. ; Since array is sorted , an element present in array must be in range defined by corner ; Probing the position with keeping uniform distribution in mind . ; Condition of target found ; If x is larger , x is in right sub array ; If x is smaller , x is in left sub array ; Driver Code ; Array of items on which search will be conducted . ; Element to be searched ; If element was found
#include <stdio.h> NEW_LINE int interpolationSearch ( int arr [ ] , int lo , int hi , int x ) { int pos ; if ( lo <= hi && x >= arr [ lo ] && x <= arr [ hi ] ) { pos = lo + ( ( ( double ) ( hi - lo ) / ( arr [ hi ] - arr [ lo ] ) ) * ( x - arr [ lo ] ) ) ; if ( arr [ pos ] == x ) return pos ; if ( arr [ pos ] < x ) return interpolationSearch ( arr , pos + 1 , hi , x ) ; if ( arr [ pos ] > x ) return interpolationSearch ( arr , lo , pos - 1 , x ) ; } return -1 ; } int main ( ) { int arr [ ] = { 10 , 12 , 13 , 16 , 18 , 19 , 20 , 21 , 22 , 23 , 24 , 33 , 35 , 42 , 47 } ; int n = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; int x = 18 ; int index = interpolationSearch ( arr , 0 , n - 1 , x ) ; if ( index != -1 ) printf ( " Element ▁ found ▁ at ▁ index ▁ % d " , index ) ; else printf ( " Element ▁ not ▁ found . " ) ; return 0 ; }
Merge Sort | C program for Merge Sort ; Merges two subarrays of arr [ ] . First subarray is arr [ l . . m ] Second subarray is arr [ m + 1. . r ] ; Find sizes of two subarrays to be merged ; create temp arrays ; Copy data to temp arrays L [ ] and R [ ] ; Merge the temp arrays Initial indexes of first and second subarrays ; Initial index of merged subarray ; Copy the remaining elements of L [ ] , if there are any ; Copy the remaining elements of R [ ] , if there are any ; l is for left index and r is right index of the sub - array of arr to be sorted ; Same as ( l + r ) / 2 , but avoids overflow for large l and h ; Sort first and second halves ; Merge the sorted halves ; Function to print an array ; Driver code
#include <stdio.h> NEW_LINE #include <stdlib.h> NEW_LINE void merge ( int arr [ ] , int l , int m , int r ) { int i , j , k ; int n1 = m - l + 1 ; int n2 = r - m ; int L [ n1 ] , R [ n2 ] ; for ( i = 0 ; i < n1 ; i ++ ) L [ i ] = arr [ l + i ] ; for ( j = 0 ; j < n2 ; j ++ ) R [ j ] = arr [ m + 1 + j ] ; i = 0 ; j = 0 ; k = l ; while ( i < n1 && j < n2 ) { if ( L [ i ] <= R [ j ] ) { arr [ k ] = L [ i ] ; i ++ ; } else { arr [ k ] = R [ j ] ; j ++ ; } k ++ ; } while ( i < n1 ) { arr [ k ] = L [ i ] ; i ++ ; k ++ ; } while ( j < n2 ) { arr [ k ] = R [ j ] ; j ++ ; k ++ ; } } void mergeSort ( int arr [ ] , int l , int r ) { if ( l < r ) { int m = l + ( r - l ) / 2 ; mergeSort ( arr , l , m ) ; mergeSort ( arr , m + 1 , r ) ; merge ( arr , l , m , r ) ; } } void printArray ( int A [ ] , int size ) { int i ; for ( i = 0 ; i < size ; i ++ ) printf ( " % d ▁ " , A [ i ] ) ; printf ( " STRNEWLINE " ) ; } int main ( ) { int arr [ ] = { 12 , 11 , 13 , 5 , 6 , 7 } ; int arr_size = sizeof ( arr ) / sizeof ( arr [ 0 ] ) ; printf ( " Given ▁ array ▁ is ▁ STRNEWLINE " ) ; printArray ( arr , arr_size ) ; mergeSort ( arr , 0 , arr_size - 1 ) ; printf ( " Sorted array is " printArray ( arr , arr_size ) ; return 0 ; }
Iterative Quick Sort | An iterative implementation of quick sort ; A utility function to swap two elements ; This function is same in both iterative and recursive ; A [ ] -- > Array to be sorted , l -- > Starting index , h -- > Ending index ; Create an auxiliary stack ; initialize top of stack ; push initial values of l and h to stack ; Keep popping from stack while is not empty ; Pop h and l ; Set pivot element at its correct position in sorted array ; If there are elements on left side of pivot , then push left side to stack ; If there are elements on right side of pivot , then push right side to stack ; A utility function to print contents of arr ; Driver program to test above functions ; Function calling
#include <stdio.h> NEW_LINE void swap ( int * a , int * b ) { int t = * a ; * a = * b ; * b = t ; } int partition ( int arr [ ] , int l , int h ) { int x = arr [ h ] ; int i = ( l - 1 ) ; for ( int j = l ; j <= h - 1 ; j ++ ) { if ( arr [ j ] <= x ) { i ++ ; swap ( & arr [ i ] , & arr [ j ] ) ; } } swap ( & arr [ i + 1 ] , & arr [ h ] ) ; return ( i + 1 ) ; } void quickSortIterative ( int arr [ ] , int l , int h ) { int stack [ h - l + 1 ] ; int top = -1 ; stack [ ++ top ] = l ; stack [ ++ top ] = h ; while ( top >= 0 ) { h = stack [ top -- ] ; l = stack [ top -- ] ; int p = partition ( arr , l , h ) ; if ( p - 1 > l ) { stack [ ++ top ] = l ; stack [ ++ top ] = p - 1 ; } if ( p + 1 < h ) { stack [ ++ top ] = p + 1 ; stack [ ++ top ] = h ; } } } void printArr ( int arr [ ] , int n ) { int i ; for ( i = 0 ; i < n ; ++ i ) printf ( " % d ▁ " , arr [ i ] ) ; } int main ( ) { int arr [ ] = { 4 , 3 , 5 , 2 , 1 , 3 , 2 , 3 } ; int n = sizeof ( arr ) / sizeof ( * arr ) ; quickSortIterative ( arr , 0 , n - 1 ) ; printArr ( arr , n ) ; return 0 ; }
Activity Selection Problem | Greedy Algo | C program for activity selection problem . The following implementation assumes that the activities are already sorted according to their finish time ; Prints a maximum set of activities that can be done by a single person , one at a time . n -- > Total number of activities s [ ] -- > An array that contains start time of all activities f [ ] -- > An array that contains finish time of all activities ; The first activity always gets selected ; Consider rest of the activities ; If this activity has start time greater than or equal to the finish time of previously selected activity , then select it ; driver program to test above function
#include <stdio.h> NEW_LINE void printMaxActivities ( int s [ ] , int f [ ] , int n ) { int i , j ; printf ( " Following ▁ activities ▁ are ▁ selected ▁ n " ) ; i = 0 ; printf ( " % d ▁ " , i ) ; for ( j = 1 ; j < n ; j ++ ) { if ( s [ j ] >= f [ i ] ) { printf ( " % d ▁ " , j ) ; i = j ; } } } int main ( ) { int s [ ] = { 1 , 3 , 0 , 5 , 8 , 5 } ; int f [ ] = { 2 , 4 , 6 , 7 , 9 , 9 } ; int n = sizeof ( s ) / sizeof ( s [ 0 ] ) ; printMaxActivities ( s , f , n ) ; return 0 ; }
Longest Common Subsequence | DP | A Naive recursive implementation of LCS problem ; Returns length of LCS for X [ 0. . m - 1 ] , Y [ 0. . n - 1 ] ; Utility function to get max of 2 integers ; Driver program to test above function
#include <bits/stdc++.h> NEW_LINE int max ( int a , int b ) ; int lcs ( char * X , char * Y , int m , int n ) { if ( m == 0 n == 0 ) return 0 ; if ( X [ m - 1 ] == Y [ n - 1 ] ) return 1 + lcs ( X , Y , m - 1 , n - 1 ) ; else return max ( lcs ( X , Y , m , n - 1 ) , lcs ( X , Y , m - 1 , n ) ) ; } int max ( int a , int b ) { return ( a > b ) ? a : b ; } int main ( ) { char X [ ] = " AGGTAB " ; char Y [ ] = " GXTXAYB " ; int m = strlen ( X ) ; int n = strlen ( Y ) ; printf ( " Length ▁ of ▁ LCS ▁ is ▁ % d " , lcs ( X , Y , m , n ) ) ; return 0 ; }
Longest Common Subsequence | DP | Dynamic Programming C implementation of LCS problem ; Returns length of LCS for X [ 0. . m - 1 ] , Y [ 0. . n - 1 ] ; Following steps build L [ m + 1 ] [ n + 1 ] in bottom up fashion . Note that L [ i ] [ j ] contains length of LCS of X [ 0. . i - 1 ] and Y [ 0. . j - 1 ] ; L [ m ] [ n ] contains length of LCS for X [ 0. . n - 1 ] and Y [ 0. . m - 1 ] ; Utility function to get max of 2 integers ; Driver program to test above function
#include <bits/stdc++.h> NEW_LINE int max ( int a , int b ) ; int lcs ( char * X , char * Y , int m , int n ) { int L [ m + 1 ] [ n + 1 ] ; int i , j ; for ( i = 0 ; i <= m ; i ++ ) { for ( j = 0 ; j <= n ; j ++ ) { if ( i == 0 j == 0 ) L [ i ] [ j ] = 0 ; else if ( X [ i - 1 ] == Y [ j - 1 ] ) L [ i ] [ j ] = L [ i - 1 ] [ j - 1 ] + 1 ; else L [ i ] [ j ] = max ( L [ i - 1 ] [ j ] , L [ i ] [ j - 1 ] ) ; } } return L [ m ] [ n ] ; } int max ( int a , int b ) { return ( a > b ) ? a : b ; } int main ( ) { char X [ ] = " AGGTAB " ; char Y [ ] = " GXTXAYB " ; int m = strlen ( X ) ; int n = strlen ( Y ) ; printf ( " Length ▁ of ▁ LCS ▁ is ▁ % d " , lcs ( X , Y , m , n ) ) ; return 0 ; }
Min Cost Path | DP | A Naive recursive implementation of MCP ( Minimum Cost Path ) problem ; A utility function that returns minimum of 3 integers ; Returns cost of minimum cost path from ( 0 , 0 ) to ( m , n ) in mat [ R ] [ C ] ; Driver program to test above functions
#include <stdio.h> NEW_LINE #include <limits.h> NEW_LINE #define R  3 NEW_LINE #define C  3 NEW_LINE int min ( int x , int y , int z ) ; int min ( int x , int y , int z ) { if ( x < y ) return ( x < z ) ? x : z ; else return ( y < z ) ? y : z ; } int minCost ( int cost [ R ] [ C ] , int m , int n ) { if ( n < 0 m < 0 ) return INT_MAX ; else if ( m == 0 && n == 0 ) return cost [ m ] [ n ] ; else return cost [ m ] [ n ] + min ( minCost ( cost , m - 1 , n - 1 ) , minCost ( cost , m - 1 , n ) , minCost ( cost , m , n - 1 ) ) ; } int main ( ) { int cost [ R ] [ C ] = { { 1 , 2 , 3 } , { 4 , 8 , 2 } , { 1 , 5 , 3 } } ; printf ( " ▁ % d ▁ " , minCost ( cost , 2 , 2 ) ) ; return 0 ; }
Min Cost Path | DP | Dynamic Programming implementation of MCP problem ; Instead of following line , we can use int tc [ m + 1 ] [ n + 1 ] or dynamically allocate memory to save space . The following line is used to keep the program simple and make it working on all compilers . ; Initialize first column of total cost ( tc ) array ; Initialize first row of tc array ; Construct rest of the tc array ; A utility function that returns minimum of 3 integers ; Driver program to test above functions
#include <stdio.h> NEW_LINE #include <limits.h> NEW_LINE #define R  3 NEW_LINE #define C  3 NEW_LINE int min ( int x , int y , int z ) ; int minCost ( int cost [ R ] [ C ] , int m , int n ) { int i , j ; int tc [ R ] [ C ] ; tc [ 0 ] [ 0 ] = cost [ 0 ] [ 0 ] ; for ( i = 1 ; i <= m ; i ++ ) tc [ i ] [ 0 ] = tc [ i - 1 ] [ 0 ] + cost [ i ] [ 0 ] ; for ( j = 1 ; j <= n ; j ++ ) tc [ 0 ] [ j ] = tc [ 0 ] [ j - 1 ] + cost [ 0 ] [ j ] ; for ( i = 1 ; i <= m ; i ++ ) for ( j = 1 ; j <= n ; j ++ ) tc [ i ] [ j ] = min ( tc [ i - 1 ] [ j - 1 ] , tc [ i - 1 ] [ j ] , tc [ i ] [ j - 1 ] ) + cost [ i ] [ j ] ; return tc [ m ] [ n ] ; } int min ( int x , int y , int z ) { if ( x < y ) return ( x < z ) ? x : z ; else return ( y < z ) ? y : z ; } int main ( ) { int cost [ R ] [ C ] = { { 1 , 2 , 3 } , { 4 , 8 , 2 } , { 1 , 5 , 3 } } ; printf ( " ▁ % d ▁ " , minCost ( cost , 2 , 2 ) ) ; return 0 ; }
0 | A Dynamic Programming based solution for 0 - 1 Knapsack problem ; A utility function that returns maximum of two integers ; Returns the maximum value that can be put in a knapsack of capacity W ; Build table K [ ] [ ] in bottom up manner ; Driver Code
#include <stdio.h> NEW_LINE int max ( int a , int b ) { return ( a > b ) ? a : b ; } int knapSack ( int W , int wt [ ] , int val [ ] , int n ) { int i , w ; int K [ n + 1 ] [ W + 1 ] ; for ( i = 0 ; i <= n ; i ++ ) { for ( w = 0 ; w <= W ; w ++ ) { if ( i == 0 w == 0 ) K [ i ] [ w ] = 0 ; else if ( wt [ i - 1 ] <= w ) K [ i ] [ w ] = max ( val [ i - 1 ] + K [ i - 1 ] [ w - wt [ i - 1 ] ] , K [ i - 1 ] [ w ] ) ; else K [ i ] [ w ] = K [ i - 1 ] [ w ] ; } } return K [ n ] [ W ] ; } int main ( ) { int val [ ] = { 60 , 100 , 120 } ; int wt [ ] = { 10 , 20 , 30 } ; int W = 50 ; int n = sizeof ( val ) / sizeof ( val [ 0 ] ) ; printf ( " % d " , knapSack ( W , wt , val , n ) ) ; return 0 ; }
Egg Dropping Puzzle | DP |  ; A utility function to get maximum of two integers ; Function to get minimum number of trials needed in worst case with n eggs and k floors ; If there are no floors , then no trials needed . OR if there is one floor , one trial needed . ; We need k trials for one egg and k floors ; Consider all droppings from 1 st floor to kth floor and return the minimum of these values plus 1. ; Driver program to test to pront printDups
#include <limits.h> NEW_LINE #include <stdio.h> NEW_LINE int max ( int a , int b ) { return ( a > b ) ? a : b ; } int eggDrop ( int n , int k ) { if ( k == 1 k == 0 ) return k ; if ( n == 1 ) return k ; int min = INT_MAX , x , res ; for ( x = 1 ; x <= k ; x ++ ) { res = max ( eggDrop ( n - 1 , x - 1 ) , eggDrop ( n , k - x ) ) ; if ( res < min ) min = res ; } return min + 1 ; } int main ( ) { int n = 2 , k = 10 ; printf ( " nMinimum ▁ number ▁ of ▁ trials ▁ in ▁ " " worst ▁ case ▁ with ▁ % d ▁ eggs ▁ and ▁ " " % d ▁ floors ▁ is ▁ % d ▁ STRNEWLINE " , n , k , eggDrop ( n , k ) ) ; return 0 ; }
Word Wrap Problem | DP | A Dynamic programming solution for Word Wrap Problem ; A utility function to print the solution ; l [ ] represents lengths of different words in input sequence . For example , l [ ] = { 3 , 2 , 2 , 5 } is for a sentence like " aaa ▁ bb ▁ cc ▁ ddddd " . n is size of l [ ] and M is line width ( maximum no . of characters that can fit in a line ) ; For simplicity , 1 extra space is used in all below arrays extras [ i ] [ j ] will have number of extra spaces if words from i to j are put in a single line ; lc [ i ] [ j ] will have cost of a line which has words from i to j ; c [ i ] will have total cost of optimal arrangement of words from 1 to i ; p [ ] is used to print the solution . ; calculate extra spaces in a single line . The value extra [ i ] [ j ] indicates extra spaces if words from word number i to j are placed in a single line ; Calculate line cost corresponding to the above calculated extra spaces . The value lc [ i ] [ j ] indicates cost of putting words from word number i to j in a single line ; Calculate minimum cost and find minimum cost arrangement . The value c [ j ] indicates optimized cost to arrange words from word number 1 to j . ; Driver program to test above functions
#include <limits.h> NEW_LINE #include <stdio.h> NEW_LINE #define INF  INT_MAX NEW_LINE int printSolution ( int p [ ] , int n ) ; int printSolution ( int p [ ] , int n ) { int k ; if ( p [ n ] == 1 ) k = 1 ; else k = printSolution ( p , p [ n ] - 1 ) + 1 ; printf ( " Line ▁ number ▁ % d : ▁ From ▁ word ▁ no . ▁ % d ▁ to ▁ % d ▁ STRNEWLINE " , k , p [ n ] , n ) ; return k ; } void solveWordWrap ( int l [ ] , int n , int M ) { int extras [ n + 1 ] [ n + 1 ] ; int lc [ n + 1 ] [ n + 1 ] ; int c [ n + 1 ] ; int p [ n + 1 ] ; int i , j ; for ( i = 1 ; i <= n ; i ++ ) { extras [ i ] [ i ] = M - l [ i - 1 ] ; for ( j = i + 1 ; j <= n ; j ++ ) extras [ i ] [ j ] = extras [ i ] [ j - 1 ] - l [ j - 1 ] - 1 ; } for ( i = 1 ; i <= n ; i ++ ) { for ( j = i ; j <= n ; j ++ ) { if ( extras [ i ] [ j ] < 0 ) lc [ i ] [ j ] = INF ; else if ( j == n && extras [ i ] [ j ] >= 0 ) lc [ i ] [ j ] = 0 ; else lc [ i ] [ j ] = extras [ i ] [ j ] * extras [ i ] [ j ] ; } } c [ 0 ] = 0 ; for ( j = 1 ; j <= n ; j ++ ) { c [ j ] = INF ; for ( i = 1 ; i <= j ; i ++ ) { if ( c [ i - 1 ] != INF && lc [ i ] [ j ] != INF && ( c [ i - 1 ] + lc [ i ] [ j ] < c [ j ] ) ) { c [ j ] = c [ i - 1 ] + lc [ i ] [ j ] ; p [ j ] = i ; } } } printSolution ( p , n ) ; } int main ( ) { int l [ ] = { 3 , 2 , 2 , 5 } ; int n = sizeof ( l ) / sizeof ( l [ 0 ] ) ; int M = 6 ; solveWordWrap ( l , n , M ) ; return 0 ; }
Optimal Binary Search Tree | DP | A naive recursive implementation of optimal binary search tree problem ; A utility function to get sum of array elements freq [ i ] to freq [ j ] ; A recursive function to calculate cost of optimal binary search tree ; Base cases no elements in this subarray ; one element in this subarray ; Get sum of freq [ i ] , freq [ i + 1 ] , ... freq [ j ] ; Initialize minimum value ; One by one consider all elements as root and recursively find cost of the BST , compare the cost with min and update min if needed ; Return minimum value ; The main function that calculates minimum cost of a Binary Search Tree . It mainly uses optCost ( ) to find the optimal cost . ; Here array keys [ ] is assumed to be sorted in increasing order . If keys [ ] is not sorted , then add code to sort keys , and rearrange freq [ ] accordingly . ; Driver program to test above functions
#include <stdio.h> NEW_LINE #include <limits.h> NEW_LINE int sum ( int freq [ ] , int i , int j ) { int s = 0 ; for ( int k = i ; k <= j ; k ++ ) s += freq [ k ] ; return s ; } int optCost ( int freq [ ] , int i , int j ) { if ( j < i ) return 0 ; if ( j == i ) return freq [ i ] ; int fsum = sum ( freq , i , j ) ; int min = INT_MAX ; for ( int r = i ; r <= j ; ++ r ) { int cost = optCost ( freq , i , r - 1 ) + optCost ( freq , r + 1 , j ) ; if ( cost < min ) min = cost ; } return min + fsum ; } int optimalSearchTree ( int keys [ ] , int freq [ ] , int n ) { return optCost ( freq , 0 , n - 1 ) ; } int main ( ) { int keys [ ] = { 10 , 12 , 20 } ; int freq [ ] = { 34 , 8 , 50 } ; int n = sizeof ( keys ) / sizeof ( keys [ 0 ] ) ; printf ( " Cost ▁ of ▁ Optimal ▁ BST ▁ is ▁ % d ▁ " , optimalSearchTree ( keys , freq , n ) ) ; return 0 ; }
Optimal Binary Search Tree | DP | Dynamic Programming code for Optimal Binary Search Tree Problem ; A utility function to get sum of array elements freq [ i ] to freq [ j ] ; A Dynamic Programming based function that calculates minimum cost of a Binary Search Tree . ; Create an auxiliary 2D matrix to store results of subproblems ; For a single key , cost is equal to frequency of the key ; Now we need to consider chains of length 2 , 3 , ... . L is chain length . ; i is row number in cost [ ] [ ] ; Get column number j from row number i and chain length L ; Try making all keys in interval keys [ i . . j ] as root ; c = cost when keys [ r ] becomes root of this subtree ; Driver program to test above functions
#include <stdio.h> NEW_LINE #include <limits.h> NEW_LINE int sum ( int freq [ ] , int i , int j ) { int s = 0 ; for ( int k = i ; k <= j ; k ++ ) s += freq [ k ] ; return s ; } int optimalSearchTree ( int keys [ ] , int freq [ ] , int n ) { int cost [ n ] [ n ] ; for ( int i = 0 ; i < n ; i ++ ) cost [ i ] [ i ] = freq [ i ] ; for ( int L = 2 ; L <= n ; L ++ ) { for ( int i = 0 ; i <= n - L + 1 ; i ++ ) { int j = i + L - 1 ; cost [ i ] [ j ] = INT_MAX ; for ( int r = i ; r <= j ; r ++ ) { int c = ( ( r > i ) ? cost [ i ] [ r - 1 ] : 0 ) + ( ( r < j ) ? cost [ r + 1 ] [ j ] : 0 ) + sum ( freq , i , j ) ; if ( c < cost [ i ] [ j ] ) cost [ i ] [ j ] = c ; } } } return cost [ 0 ] [ n - 1 ] ; } int main ( ) { int keys [ ] = { 10 , 12 , 20 } ; int freq [ ] = { 34 , 8 , 50 } ; int n = sizeof ( keys ) / sizeof ( keys [ 0 ] ) ; printf ( " Cost ▁ of ▁ Optimal ▁ BST ▁ is ▁ % d ▁ " , optimalSearchTree ( keys , freq , n ) ) ; return 0 ; }