0/1 Knapsack DP #include #include void knapsack(int n, int m, int profits[], int weights[]) { int v[n+1][m+1]; int result[n]; for(int i=0; i<=m; i++) v[0][i] = 0; for(int i=0; i<=n; i++) v[i][0] = 0; for(int i=1; i<=n; i++) { for(int w=1; w<=m; w++) { if(w - weights[i-1] <0) { v[i][w] = v[i-1][w]; } else { int val1 = v[i-1][w]; int val2 = v[i-1][w-weights[i-1]] + profits[i-1]; v[i][w] = (val1 > val2) ? val1 : val2; } } } int k=n, j=m; while (k>0 && j>0) { if(v[k][j] != v[k-1][j]) { result[k-1] = 1; j = j-weights[k-1]; } else { result[k-1] = 0; } k--; } for(int i=0; i #include char S1[20], S2[20]; int i, j, m, n, LCS_table[20][20]; int max(int a, int b) { return (a>b) ? a: b; } void lcsAlgo() { printf("Reached\n"); m = strlen(S1); n = strlen(S2); for(i = 0; i<=m; i++) LCS_table[i][0] = 0; for(i=0; i<=n; i++) LCS_table[0][i] = 0; for(i=1; i<=m; i++) { for(j=1; j<=n; j++) { if(S1[i-1] == S2[j-1]) LCS_table[i][j] = LCS_table[i-1][j-1] + 1; else LCS_table[i][j] = max(LCS_table[i-1][j], LCS_table[i][j-1]); } } int index = LCS_table[m][n]; char LCSAlgo[index+1]; LCSAlgo[index] = '\0'; printf("%d\n", index); i=m, j=n; while(i>0 && j>0) { if(S1[i-1] == S2[j-1]) { LCSAlgo[index-1] = S1[i-1]; i--; j--; index--; } else { //max(LCS_table[i-1][j], LCS_table[i][j-1]); if(LCS_table[i - 1][j] > LCS_table[i][j - 1]) i--; else j--; } } printf("%d\n", index); printf("The LCS is %s", LCSAlgo); } int main() { scanf("%[^\n]s", &S1); scanf("%s", &S2); lcsAlgo(); } Matrix Chain #include #include int MatrixChainMultiplication(int arr[], int n) { int minMul[n][n]; for(int i=1; i #include #define min(a, b) ((a) < (b) ? (a) : (b)) int fun(int a*2+*4+, int t*2+*4+, int cl, int cs, int x1, int x2, int n); int main() , int n = 4; // number of stations int a*2+*4+ = , , 4, 5, 3, 2 -, , 2, 10, 1, 4 - -; int t*2+*4+ = , , 0, 7, 4, 5 -, , 0, 9, 2, 8 - -; int e1 = 10; int e2 = 12; int x1 = 18; int x2 = 7; // entry from 1st line int x = fun(a, t, 0, 0, x1, x2, n) + e1 + a*0+*0+; // entry from 2nd line int y = fun(a, t, 1, 0, x1, x2, n) + e2 + a*1+*0+; printf("%d\n", min(x, y)); return 0; - int fun(int a*2+*4+, int t*2+*4+, int cl, int cs, int x1, int x2, int n) , // base case if (cs == n - 1) , if (cl == 0) , // exiting from (current) line =0 return x1; - else , // exiting from line 2 return x2; - - // continue on same line int same = fun(a, t, cl, cs + 1, x1, x2, n) + a*cl+*cs + 1+; // continue on different line int diff = fun(a, t, !cl, cs + 1, x1, x2, n) + a*!cl+*cs + 1+ + t*cl+*cs + 1+; return min(same, diff); - Max Sub Array Sum #include #include int *maxSubArraySum(int nums[], int start, int end, int *max_sum, int *subarray_start, int *subarray_end) { if(start == end) { *max_sum = nums[start]; *subarray_start = start; *subarray_end = start; return &nums[start]; } int mid= (start+end)/2; int left_subarray = maxSubArraySum(nums, start, mid, &max_sum, &subarray_start, &subarray_end); int right_subarray = maxSubArraySum(nums, mid+1, end, &max_sum, &subarray_start, &subarray_end); int left_sum=INT_MIN; int curr_sum = 0; int cross_start = mid; for(int i=mid; i>=start; i--) { curr_sum += nums[i]; if(curr_sum >= left_sum) { left_sum = curr_sum; cross_start = i; } } int right_sum=INT_MIN; curr_sum=0; int cross_end = mid+1; for(int i=mid+1; i= right_sum) { right_sum = curr_sum; cross_end = i; } } } int main() { int n; scanf("%d", &n); int nums[n]; for(int i=0; i #include // Function to find the sum of larger // numbers represented as a string char* findSum(const char* str1, const char* str2) , // Before proceeding further, make // sure length of str2 is larger if (strlen(str1) > strlen(str2)) , const char* temp = str1; str1 = str2; str2 = temp; - // Stores the result static char str*1000+; // Calculate length of both strings int n1 = strlen(str1); int n2 = strlen(str2); // Reverse both of strings strrev(str1); strrev(str2); int carry = 0; for (int i = 0; i < n1; i++) , // Find the sum of the current // digits and carry int sum = ((str1*i+ - '0') + (str2*i+ - '0') + carry); str*i+ = (sum % 10) + '0'; // Calculate carry for next step carry = sum / 10; - // Add remaining digits of larger number for (int i = n1; i < n2; i++) , int sum = ((str2*i+ - '0') + carry); str*i+ = (sum % 10) + '0'; carry = sum / 10; - // Add remaining carry if (carry) str*n2+ = carry + '0'; else str*n2+ = '\0'; // Reverse resultant string strrev(str); return str; - // Function to find difference of larger // numbers represented as strings char* findDiff(const char* str1, const char* str2) , // Stores the result of difference static char str*1000+; // Calculate length of both strings int n1 = strlen(str1), n2 = strlen(str2); // Reverse both of strings strrev(str1); strrev(str2); int carry = 0; // Run loop till small string length // and subtract digit of str1 to str2 for (int i = 0; i < n2; i++) , // Compute difference of the // current digits int sub = ((str1*i+ - '0') - (str2*i+ - '0') - carry); // If subtraction < 0 then add 10 // into sub and take carry as 1 if (sub < 0) , sub = sub + 10; carry = 1; - else carry = 0; str*i+ = sub + '0'; - // Subtract the remaining digits of // larger number for (int i = n2; i < n1; i++) , int sub = ((str1*i+ - '0') - carry); // If the sub value is -ve, // then make it positive if (sub < 0) , sub = sub + 10; carry = 1; - else carry = 0; str*i+ = sub + '0'; - // Reverse resultant string strrev(str); // Return answer return str; - // Function to remove all leading 0s // from a given string char* removeLeadingZeros(char* str) , // Index to store the position of // the first non-zero digit int i, len = strlen(str); for (i = 0; i < len - 1; i++) , if (str*i+ != '0') break; - // Shift all non-zero digits to the // beginning of the string for (int j = 0; j < len - i; j++) str*j+ = str*i + j+; // Null terminate the string str*len - i+ = '\0'; return str; - // Function to multiply two numbers // using Karatsuba algorithm char* multiply(const char* A, const char* B) , const char* str1 = A; const char* str2 = B; if (strlen(A) > strlen(B)) , str1 = B; str2 = A; - // Make both numbers to have // same digits int n1 = strlen(str1), n2 = strlen(str2); while (n2 > n1) , n1++; - // Base case if (n1 == 1) , // If the length of strings is 1, // then return their product int ans = atoi(str1) * atoi(str2); sprintf(str1, "%d", ans); return str1; - // Add zeros in the beginning of // the strings when length is odd if (n1 % 2 == 1) , n1++; char temp*1000+; strcpy(temp, str1); strcpy(str1, "0"); strcat(str1, temp); strcpy(temp, str2); strcpy(str2, "0"); strcat(str2, temp); - char Al*1000+, Ar*1000+, Bl*1000+, Br*1000+; // Find the values of Al, Ar, // Bl, and Br. for (int i = 0; i < n1 / 2; ++i) , Al*i+ = str1*i+; Bl*i+ = str2*i+; Ar*i+ = str1*n1 / 2 + i+; Br*i+ = str2*n1 / 2 + i+; - Al*n1 / 2+ = '\0'; Ar*n1 / 2+ = '\0'; Bl*n1 / 2+ = '\0'; Br*n1 / 2+ = '\0'; // Recursively call the function // to compute smaller product // Stores the value of Al * Bl char* p = multiply(Al, Bl); // Stores the value of Ar * Br char* q = multiply(Ar, Br); // Stores value of ((Al + Ar)*(Bl + Br) // - Al*Bl - Ar*Br) char* r = findDiff( findSum(Al, Ar), findSum(Bl, Br)); // Multiply p by 10^n for (int i = 0; i < n1; ++i) strcat(p, "0"); // Multiply s by 10^(n/2) for (int i = 0; i < n1 / 2; ++i) strcat(r, "0"); // Calculate final answer p + r + s char* ans = findSum(p, findSum(q, r)); Subset Sum #include #include #include bool foundSolution = false; int weights[10]; int solution[10]; int n; int m; void subsetSum(int currentSum, int index) { if(index == n) { if(currentSum == m) { foundSolution = true; for(int i=0; i #include // Structure for an item which stores weight and // corresponding value of Item struct Item , int profit, weight; -; // Comparison function to sort Item // according to profit/weight ratio static int cmp(const void* a, const void* b) , double r1 = (double)((struct Item*)b)->profit / (double)((struct Item*)b)->weight; double r2 = (double)((struct Item*)a)->profit / (double)((struct Item*)a)->weight; if (r1 > r2) return 1; else if (r1 < r2) return -1; return 0; - // Main greedy function to solve problem double fractionalKnapsack(int W, struct Item arr*+, int N) , // Sorting Item on basis of ratio qsort(arr, N, sizeof(struct Item), cmp); double finalvalue = 0.0; // Looping through all items for (int i = 0; i < N; i++) , // If adding Item won't overflow, // add it completely if (arr*i+.weight <= W) , W -= arr*i+.weight; finalvalue += arr*i+.profit; - // If we can't add current Item, // add fractional part of it else , finalvalue += arr*i+.profit * ((double)W / (double)arr*i+.weight); break; - - // Returning final value return finalvalue; - // Driver code int main() , int W = 50; struct Item arr*+ = , , 60, 10 -, , 100, 20 -, , 120, 30 - -; int N = sizeof(arr) / sizeof(arr*0+); // Function call printf("%.2lf\n", fractionalKnapsack(W, arr, N)); return 0; - N QUENS CODE : // C program to solve N Queen Problem using backtracking #define N 4 #include #include // A utility function to print solution void printSolution(int board*N+*N+) , for (int i = 0; i < N; i++) , for (int j = 0; j < N; j++) , if(board*i+*j+) printf("Q "); else printf(". "); - printf("\n"); - - // A utility function to check if a queen can // be placed on board*row+*col+. Note that this // function is called when "col" queens are // already placed in columns from 0 to col -1. // So we need to check only left side for // attacking queens bool isSafe(int board*N+*N+, int row, int col) , int i, j; // Check this row on left side for (i = 0; i < col; i++) if (board*row+*i+) return false; // Check upper diagonal on left side for (i = row, j = col; i >= 0 && j >= 0; i--, j--) if (board*i+*j+) return false; // Check lower diagonal on left side for (i = row, j = col; j >= 0 && i < N; i++, j--) if (board*i+*j+) return false; return true; - // A recursive utility function to solve N // Queen problem bool solveNQUtil(int board*N+*N+, int col) , // Base case: If all queens are placed // then return true if (col >= N) return true; // Consider this column and try placing // this queen in all rows one by one for (int i = 0; i < N; i++) , // Check if the queen can be placed on // board*i+*col+ if (isSafe(board, i, col)) , // Place this queen in board*i+*col+ board*i+*col+ = 1; // Recur to place rest of the queens if (solveNQUtil(board, col + 1)) return true; // If placing queen in board*i+*col+ // doesn't lead to a solution, then // remove queen from board*i+*col+ board*i+*col+ = 0; // BACKTRACK - - // If the queen cannot be placed in any row in // this column col then return false return false; - // This function solves the N Queen problem using // Backtracking. It mainly uses solveNQUtil() to // solve the problem. It returns false if queens // cannot be placed, otherwise, return true and // prints placement of queens in the form of 1s. // Please note that there may be more than one // solutions, this function prints one of the // feasible solutions. bool solveNQ() , int board*N+*N+ = , , 0, 0, 0, 0 -, , 0, 0, 0, 0 -, , 0, 0, 0, 0 -, , 0, 0, 0, 0 - -; if (solveNQUtil(board, 0) == false) , printf("Solution does not exist"); return false; - printSolution(board); return true; - // Driver program to test above function int main() , solveNQ(); return 0; - // This code is contributed by Aditya Kumar (adityakumar129) RANDOMIZED QUICK SORT #include #include #include int partition(int arr*+, int low, int high) , int pivot = arr*low+; int i = low - 1, j = high + 1; while (1) , do , i++; - while (arr*i+ < pivot); do , j--; - while (arr*j+ > pivot); if (i >= j) return j; int temp = arr*i+; arr*i+ = arr*j+; arr*j+ = temp; - - int partition_r(int arr*+, int low, int high) , srand(time(0)); int random = low + rand() % (high - low); int temp = arr*random+; arr*random+ = arr*low+; arr*low+ = temp; return partition(arr, low, high); - void quickSort(int arr*+, int low, int high) , if (low < high) , int pi = partition_r(arr, low, high); quickSort(arr, low, pi); quickSort(arr, pi + 1, high); - - void printArray(int arr*+, int n) , for (int i = 0; i < n; i++) printf("%d ", arr*i+); printf("\n"); - int main() , int arr*+ = , 10, 7, 8, 9, 1, 5 -; int n = sizeof(arr) / sizeof(arr*0+); quickSort(arr, 0, n - 1); printf("Sorted array: \n"); printArray(arr, n); return 0; -