0/1 Knapsack DP
#include <stdio.h>
#include <limits.h>
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<n; i++)
printf("%d ", result[i]);
}
int main()
{
int n; //No. of objects
scanf("%d", &n);
int m; //Max capacity
scanf("%d", &m);
int profits[n], weights[n];
for(int i=0; i<n; i++)
scanf("%d", &profits[i]);
for(int i=0; i<n; i++)
scanf("%d", &weights[i]);
knapsack(n, m, profits, weights);
return 0;
}
LCS
#include <stdio.h>
#include <string.h>
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 <stdio.h>
#include <limits.h>
int MatrixChainMultiplication(int arr[], int n)
{
int minMul[n][n];
for(int i=1; i<n; i++)
minMul[i][i] = 0;
int j, q=0;
for(int L=2; L<n; L++) {
for(int i=1; i<n-L+1; i++) {
j = i+L-1;
minMul[i][j] = INT_MAX;
for(int k=i; k<=j; k++) {
q = minMul[i][k] + minMul[k+1][j] + arr[i-1]*arr[k]*arr[j];
if(q<minMul[i][j])
minMul[i][j]=q;
}
}
}
return minMul[1][n-1];
}
int main()
{
int n;
scanf("%d", &n);
int arrS[n][2];
int arr[n+1];
for(int i=0; i<n; i++) {
scanf("%d %d",&arrS[i][0], &arrS[i][1]);
}
arr[0] = arrS[0][0];
for(int i=1; i<=n; i++)
arr[i] = arrS[i-1][1];
printf("Minimum number of multiplications required for matrix
multiplication is %d\n", MatrixChainMultiplication(arr, n+1));
getchar();
return 0;
}
Assembly line scheduling
Code:
#include <stdio.h>
#include <stdlib.h>
#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 <stdio.h>
#include <limits.h>
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<end; i++) {
curr_sum+=nums[i];
if(curr_sum>= right_sum) {
right_sum = curr_sum;
cross_end = i;
}
}
}
int main()
{
int n;
scanf("%d", &n);
int nums[n];
for(int i=0; i<n; i++)
scanf("%d ", &nums[i]);
int max_sum, subarray_start, subarray_end;
int *max_subarray =
maxSubArraySum(nums, 0, n-1, &max_sum, &subarray_start,
&subarray_end);
printf("The sub array [");
for(int i=subarray_start; i<=subarray_end; i++) {
if(i==subarray_end)
printf("%d", nums[i]);
else
printf("%d ", nums[i]);
}
printf("] has the largest sum %d", max_sum);
return 0;
}
KARATSUBA ALGORITHM
#include <stdio.h>
#include <string.h>
// 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 <stdio.h>
#include <limits.h>
#include <stdbool.h>
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<n; i++)
printf("%d ", solution[i]);
printf("\n");
}
return;
}
solution[index] = 1;
subsetSum(currentSum + weights[index], index+1);
solution[index] = 0;
subsetSum(currentSum, index + 1);
}
int main()
{
//int n, m;
scanf("%d", &n);
scanf("%d", &m);
//int weights[n];
for(int i=0; i<n; i++)
scanf("%d", &weights[i]);
subsetSum(0, 0);
if(!foundSolution)
printf("No solution found.\n");
return 0;
}
FRACTIONAL KNAPSACK GREEDY APPROACH
#include <stdio.h>
#include <stdlib.h>
// 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 <stdbool.h>
#include <stdio.h>
// 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 <stdio.h>
#include <stdlib.h>
#include <time.h>
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;
-