NumN: 1, 2, 3, 4, 5 Let's say 3 & 4 are missing (x and y) ListN : 1, 2, 5 x + y = 7 -- (1) // SumOfNumN - SumOfListN x^2 + y^2 = 25 -- (2) // (sumOf the squares in NumN) - (sumOf the squares in ListN) Substitute (1) in (2), x^2 + (7 - x)^2 = 25 Roots: 3, 4 which are the missing numbers
Code :
#include<iostream> using namespace std; void Random_List(int n) { int a[10000]; for(int i=1;i<=n;i++) { a[i] = i; } cout<<"randlist("<<n<<") :"; for(int i=n;i>=1;i--) { int j = rand() % i + 1; int swap = a[j]; a[j] = a[i]; a[i] = swap; } for(int i=1;i<=n;i++) { cout<<" "<<a[i]; } } int main() { int n = 0; cout<<"n=";cin>>n; Random_List(n); cout<<endl; return 0; }
Code :
#include <stdio.h> #include <string.h> #include <stdlib.h> struct node_t { int c; struct node_t *next; }; struct node_t *create_node(char c) { struct node_t* node = (struct node_t*) malloc(sizeof(struct node_t)); node->c = c; node->next = NULL; return node; } void print(struct node_t *head) { struct node_t *p = head; while (p != NULL) { printf("%c", p->c); p = p->next; } printf("\n"); } int main() { int x; struct node_t *head = NULL; struct node_t *node; while ( (x = getchar()) != 10 ) { node = create_node(x); node->next = head; head = node; } print(head); return 0;
}
Code :
#include<iostream> using namespace std; /* How to check if a binary tree is a binary search tree */ typedef struct node_s { int value; struct node_s *left; struct node_s *right; }BSTNode, *BSTree; int tree_search(BSTree T, int value, BSTNode **p, BSTNode *f) { if(!T) { *p = f; return 0; } else { if(T->value == value) { *p = T; return 1; } else if(value < T->value) { return tree_search(T->left, value, p, T); } else { return tree_search(T->right, value, p, T); } } } int tree_insert(BSTree *T, int value) { BSTNode *p = NULL; if(!tree_search(*T, value, &p, NULL)) { BSTNode *s = (BSTNode*)malloc(sizeof(BSTNode)); s->value = value; s->left = NULL; s->right = NULL; if(!(*T)) *T = s; else if(value < p->value) p->left = s; else p->right = s; return 1; } return 0; } bool is_bst = true; void check_tree_is_bst(BSTree T) { if(T) { check_tree_is_bst(T->left); if(T->left) { if(T->left->value > T->value) { is_bst = false; return; } } if(T->right) { if(T->right->value < T->value) { is_bst = false; return; } } check_tree_is_bst(T->right); } } int main(int argc, char *argv[]) { int a[] = {5, 9, 13, 4, 6, 7, 34, 12, 25, 16}; int len = sizeof(a) / sizeof(int); int i; BSTree T = NULL; for(i = 0; i < len; i++) tree_insert(&T, a[i]); check_tree_is_bst(T); if(is_bst) cout << "yes" << endl; else cout << "no" << endl; return 0; }
#include <stdio.h>
#define N 5
#define M 3
int min(int a, int b)
{
return a < b ? a : b;
}
void spiral(int mat[M][N])
{
int n = N, d = +1, i = 0, j = -1;
int s, segl = N + M - 1;
for (s = 0; s < min(N, M); s++, d = -d, segl -= 2, n—) {
int c, *ip = &j;
for (c = 0; c < segl; c++) {
if (c == n)
ip = &i;
*ip += d;
printf(“%3d”, mat[i][j]);
}
}
}
Code :
maxsofar = 0 maxendinghere = 0 for i = [0, n)] /* invariant: maxendinghere and maxsofar are accurate are accurate for x[0..i-1] */ maxendinghere = max(maxendinghere + x[i], 0) maxsofar = max(maxsofar, maxendinghere)
Code :
void mergeArray(int* arr1, int n1, int* arr2, int n2) { int i, j, k; i = n1-1; j = n2-1; k = n1+n2-1; while(i>=0 && j>=0 && k >= 0) { if(arr1[i] >= arr2[j]) { arr1[k] = arr1[i]; --k; --i; } else { arr1[k] = arr2[j]; --k; --j; } } while(i>=0) { arr1[k--] = arr1[i--]; } while(j>=0) { arr1[k--] = arr2[i--]; } }
Code :
main()
{
char s[]=”my name is anthony”;
revword(s);
printf(“\nafter rev : %s”,s);
return 0;
}void revword(char *str)
{
char *end,*x,*y,i;
for(end=str;*end;end++){}
/*reverse complete sentence */
rev(str,end-1);
/*reverse word in the sentence */
x=str; y=str;
while(x++<end)
{if(*x==” || *x==’/0’)
{
rev(y,x-1);
y=x+1;
}
}
}void rev(char *r,char *l)
{
char temp;
while(r<l)
{
temp=*r;
*r=*l;
*l=temp;
r++;
l—;
}
}
Code:
Sort the array using merg sort O(nlogn)
Search the Array : This one has the time complexity O(n3)
<script>
var arr=[“12”,”10”,”9”,”7”,”3”,”1”,”0”];
var arr=[“15”,”10”,”7”,”2”,”2”,”2”,”2”];
var maxSum=maxItemSum(arr);function maxItemSum(a)
{
var len=a.length;
var i=0,j=0,k=0;
var flag=’notfound’;
for(i=0;i<len-2;i++)
{
for(j=i+1;j<len-1;j++)
{
for(k=j+1;k<len;k++)
{
if(parseInt(a[i])==parseInt(a[j])+parseInt(a[k]))
{
document.write(a[i]+’equals’+a[j]+’plus’+a[k]+’<br/>’);
flag=’found’;
break;
}
}
if(flag==’found’)
break;
}
if(flag==’found’)
break;
}
}
</script>
A better Optimized version :
<script>
var arr=[“12”,”10”,”9”,”7”,”3”,”2”,”0”];
/*var arr=[“15”,”10”,”7”,”2”,”2”,”2”,”2”];*/
var maxSum=maxItemSum(arr);
document.write(maxSum);
function maxItemSum(a)
{
var lenF=a.length;
var i=0;
var j=0;
var flag=’notfound’;
for(i=0;i<lenF;i++)
{
j=i+1;
var len=a.length;
while(j<(len-1) && flag==’notfound’)
{
if(parseInt(a[i])==parseInt(a[j])+parseInt(a[len-1]))
{
flag=’found’;
break;
}
else if(parseInt(a[i])>parseInt(a[j])+ parseInt(a[len-1]))
{
len—;}
else if(parseInt(a[i])<parseInt(a[j])+parseInt(a[len-1]))
{
j++;
}
if(flag==’found’)
break;
}
if(flag!=’found’)
return flag;
else
return a[i];
}
}
</script>
Code:
<script>
var str1=”aBcd”;
var str2=”ABC”;
var result=lexicographicComparison(str1,str2);
if(result==1)
{
document.write(‘str1 is greater’);
}
else if(result==-1)
{
document.write(‘str2 is greater’);
}
else if(result==0)
{
document.write(‘str1 is equal to str2’);
}
function lexicographicComparison(str1,str2)
{
var len1=str1.length;
var len2=str2.length;
var sum1=0;
var sum2=0;
var i=0;
while(i<len1)
{
if(str1.charCodeAt(i)<97)
sum1+=str1.charCodeAt(i)+32;
else
sum1+=str1.charCodeAt(i);
i++;
}
i=0;
while(i<len2)
{
if(str2.charCodeAt(i)<97)
sum2+=str2.charCodeAt(i)+32;
else
sum2+=str2.charCodeAt(i);
i++;
}
document.write(sum1);
document.write(‘<br/>’+sum2+’<br/>’);
if(sum1>sum2)
return 1;
else if(sum1<sum2)
return -1;
else if(sum1 ==sum2)
return 0;
}
</script>
