CareerCup Chapter 9 Sorting and Searching
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9.1 You are given two sorted arrays, A and B, and A has a large enough buffer at the end to hold B. Write a method to merge B into A in sorted order.
A has enough buffer at the end to hold B, we can merge two arrays from end to start index, like merge two arrays in merge sort algorithm.
void mergeTwoArray(int A[],int m,int B[],int n){
int k=m+n-1;
m--;n--;
while(m>=0&&n>=0){
if(A[m]>=B[n])A[k--]=A[m--];
else A[k--]=B[n--];
}
while(m>=0)A[k--]=A[m--];
while(n>=0)A[k--]=B[n--];
}
9.2 Write a method to sort an array of strings so that all the anagrams are next to each other.
Based on basic sort function, we define a cmp function to decide how to sort. Here, we compare theirs anagrams' state.
int cmp(string s1,string s2){
sort(s1.begin(),s1.end());
sort(s2.begin(),s2.end());
if(s1 else return 0;
}
void sortStrings(vector &strs){
sort(strs.begin(),strs.end(),cmp);
}
**9.3 Given a sorted array of n integers that has been rotated an unknown number of times, give an O(log n) algorithm that finds an element in the array. You may assume that the array was originally sorted in increasing order. EXAMPLE: Input: find 5 in array (15 16 19 20 25 1 3 4 5 7 10 14)
Output: 8 (the index of 5 in the array)
like binary search, we declare left/mid/right variables to represent current array part, because the array is rotated, we will have 8 states as following: k is the goal data.
k ka[right], in this condition, left=mid+1;
ka[left],k ka[left],k>a[right], in this condition, right = mid-1;
k>a[mid],k k>a[mid],ka[right], in this condition, right=mid-1;
k>a[mid],k>a[left],k k>a[mid],k>a[left],k>a[right], in this condition, left=mid+1 || right=mid-1;
Method2: not only compare target with a[left/mid/right], we compare a[left]/a[mid]/a[right] to decide which part has rotated part.
k kleft, in this condition, right=mid-1
kmid, in this condition, right=mid-1
k>mid,left k>mid,left>mid,k>right, in this condition, right=mid-1
k>mid,left>mid,k int search(int A[],int n,int target){
int left=0,right=n-1,mid;
while(left<=right){
mid=left+(right-left)/2;
if(A[mid]==target)return mid;
else if(target if(A[left]<=A[mid]){
if(target else right=mid-1;
}else right=mid-1;
}else{
if(A[left]>A[mid]){
if(target<=A[right])left=mid+1;
else right=mid-1;
}else left=mid+1;
}
}
return -1;
}
9.4 If you have a 2 GB file with one string per line, which sorting algorithm would you use to sort the file and why?
external sort. Divide to N parts, sort each part, then N-way merge.
9.5 Given a sorted array of strings which is interspersed with empty strings, write a method to find the location of a given string. Example: find “ball” in [“at”, “”, “”, “”, “ball”, “”, “”, “car”, “”, “”, “dad”, “”, “”] will return 4
Example: find “ballcar” in [“at”, “”, “”, “”, “”, “ball”, “car”, “”, “”, “dad”, “”, “”] will return -1
we do it based on binary search, first, we traverse the front part and the back part to make true left index and right index pointing to a no-empty string. Then we find mid index, if strs[mid]=="", we traverse mid to right then to left to find the first no-empty string, this index is the mid index and do it like binary search.
int searchStrs(vector strs,string target){
if(target==""||strs.size()==0)return -1;
int left=0,right=strs.size()-1,mid;
while(left<=right){
while(strs[left]==""&&left while(strs[right]==""&&right>left)right--;
mid=left+(right-left)/2;
if(strs[left]==target)return left;
if(strs[right]==target)return right;
if(strs[mid]==""){
for(int i=mid+1;i if(i!=right)mid=i;
else{
for(int i=mid-1;i>left;i--){if(strs[i]!="")break;}
if(i!=left)mid=i
else return -1;
}
}
if(strs[mid]==target)return mid;
else if(strs[mid] else if(strs[mid]>target)right=mid-1;
}
}
return -1;
}
9.6 Given a matrix in which each row and each column is sorted, write a method to find an element in it.
the matrix is sorted in rows and columns, let's assume it is a m*n matrix, we begin at matrix[0][n-1], if target< current, we move down, if target>current, we move left, if(target ==current) we find it, if we move out of the matrix, the element isn't in it.
time complexity is O(n+m).
int rowAndColumn(vector> matrix,int target){
int m=matrix.size();if(m==0)return -1;
int n=matrix[0].size();if(n==0)return -1;
int x=0,y=n-1;
while(x=0){
if(matrix[x][y]==target)return x*n+y;
else if(matrix[x][y]>target)y--;
else if(matrix[x][y] }
return -1;
}
9.7 A circus is designing a tower routine consisting of people standing atop one another’s shoulders. For practical and aesthetic reasons, each person must be both shorter and lighter than the person below him or her. Given the heights and weights of each person in the circus, write a method to compute the largest possible number of people in such a tower. EXAMPLE: Input (ht, wt): (65, 100) (70, 150) (56, 90) (75, 190) (60, 95) (68, 110) Output: The longest tower is length 6 and includes from top to bottom: (56, 90) (60,95) (65,100) (68,110) (70,150) (75,190)
we first sort all people by their height, then find the longest ascending sequence of their wight.
struct People{
int height;
int weight;
};
int cmp(People* p1,People* p2){
if(p1->heightheight)return 1;
else return 0;
}
int largestNum(vector peos){
if(peos.size()==0)return 0;
sort(peos.begin(),peos.end(),cmp);
int *f = new int[peos.size()];
int maxNum=0;
for(int i=0;i f[i]=1;
for(int j=0;j if(peos[j]->weightweight){
f[i]=max(f[i],f[j]+1);
}
}
maxNum=(maxNum>f[i]?maxNum:f[i]);
}
return maxNum;
}
A has enough buffer at the end to hold B, we can merge two arrays from end to start index, like merge two arrays in merge sort algorithm.
void mergeTwoArray(int A[],int m,int B[],int n){
int k=m+n-1;
m--;n--;
while(m>=0&&n>=0){
if(A[m]>=B[n])A[k--]=A[m--];
else A[k--]=B[n--];
}
while(m>=0)A[k--]=A[m--];
while(n>=0)A[k--]=B[n--];
}
9.2 Write a method to sort an array of strings so that all the anagrams are next to each other.
Based on basic sort function, we define a cmp function to decide how to sort. Here, we compare theirs anagrams' state.
int cmp(string s1,string s2){
sort(s1.begin(),s1.end());
sort(s2.begin(),s2.end());
if(s1
}
void sortStrings(vector
sort(strs.begin(),strs.end(),cmp);
}
**9.3 Given a sorted array of n integers that has been rotated an unknown number of times, give an O(log n) algorithm that finds an element in the array. You may assume that the array was originally sorted in increasing order. EXAMPLE: Input: find 5 in array (15 16 19 20 25 1 3 4 5 7 10 14)
Output: 8 (the index of 5 in the array)
like binary search, we declare left/mid/right variables to represent current array part, because the array is rotated, we will have 8 states as following: k is the goal data.
k
k
k>a[mid],k
k>a[mid],k>a[left],k
int findIndex(int a[],int left,int right,int k){
if(left>right)return -1;
int mid=left+(right-left)/2;
if(k==a[mid])return mid;
else if(k==a[left])return left;
else if(k==a[right])return right;
else if(k<a[mid]){
if(k<a[left]){
if(k<a[right]){
int temp=findIndex(a,left,mid-1,k);
if(temp!=-1)return temp;
return findIndex(a,mid+1,right,k);
}else{
return findIndex(a,mid+1,right,k);
}
}else if(k>a[left]){
return findIndex(a,left,mid-1,k);
}
}else if(k>a[mid]){
if(k>right){
if(k>left){
int temp=findIndex(a,left,mid-1,k);
if(temp!=-1)return temp;
return findIndex(a,mid+1,right,k);
}else{
return findIndex(a,left,mid-1,k);
}
}else{
return findIndex(a,mid+1,right,k);
}
}
return -1;
}
int search(int A[], int n, int target) {
return findIndex(A,0,n-1,target);
}Method2: not only compare target with a[left/mid/right], we compare a[left]/a[mid]/a[right] to decide which part has rotated part.
k
k
k>mid,left
k>mid,left>mid,k
int left=0,right=n-1,mid;
while(left<=right){
mid=left+(right-left)/2;
if(A[mid]==target)return mid;
else if(target
if(target
}else right=mid-1;
}else{
if(A[left]>A[mid]){
if(target<=A[right])left=mid+1;
else right=mid-1;
}else left=mid+1;
}
}
return -1;
}
9.4 If you have a 2 GB file with one string per line, which sorting algorithm would you use to sort the file and why?
external sort. Divide to N parts, sort each part, then N-way merge.
9.5 Given a sorted array of strings which is interspersed with empty strings, write a method to find the location of a given string. Example: find “ball” in [“at”, “”, “”, “”, “ball”, “”, “”, “car”, “”, “”, “dad”, “”, “”] will return 4
Example: find “ballcar” in [“at”, “”, “”, “”, “”, “ball”, “car”, “”, “”, “dad”, “”, “”] will return -1
we do it based on binary search, first, we traverse the front part and the back part to make true left index and right index pointing to a no-empty string. Then we find mid index, if strs[mid]=="", we traverse mid to right then to left to find the first no-empty string, this index is the mid index and do it like binary search.
int searchStrs(vector
if(target==""||strs.size()==0)return -1;
int left=0,right=strs.size()-1,mid;
while(left<=right){
while(strs[left]==""&&left
mid=left+(right-left)/2;
if(strs[left]==target)return left;
if(strs[right]==target)return right;
if(strs[mid]==""){
for(int i=mid+1;i
else{
for(int i=mid-1;i>left;i--){if(strs[i]!="")break;}
if(i!=left)mid=i
else return -1;
}
}
if(strs[mid]==target)return mid;
else if(strs[mid]
}
}
return -1;
}
9.6 Given a matrix in which each row and each column is sorted, write a method to find an element in it.
the matrix is sorted in rows and columns, let's assume it is a m*n matrix, we begin at matrix[0][n-1], if target< current, we move down, if target>current, we move left, if(target ==current) we find it, if we move out of the matrix, the element isn't in it.
time complexity is O(n+m).
int rowAndColumn(vector
int m=matrix.size();if(m==0)return -1;
int n=matrix[0].size();if(n==0)return -1;
int x=0,y=n-1;
while(x
if(matrix[x][y]==target)return x*n+y;
else if(matrix[x][y]>target)y--;
else if(matrix[x][y]
return -1;
}
9.7 A circus is designing a tower routine consisting of people standing atop one another’s shoulders. For practical and aesthetic reasons, each person must be both shorter and lighter than the person below him or her. Given the heights and weights of each person in the circus, write a method to compute the largest possible number of people in such a tower. EXAMPLE: Input (ht, wt): (65, 100) (70, 150) (56, 90) (75, 190) (60, 95) (68, 110) Output: The longest tower is length 6 and includes from top to bottom: (56, 90) (60,95) (65,100) (68,110) (70,150) (75,190)
we first sort all people by their height, then find the longest ascending sequence of their wight.
struct People{
int height;
int weight;
};
int cmp(People* p1,People* p2){
if(p1->height
else return 0;
}
int largestNum(vector
if(peos.size()==0)return 0;
sort(peos.begin(),peos.end(),cmp);
int *f = new int[peos.size()];
int maxNum=0;
for(int i=0;i
for(int j=0;j
f[i]=max(f[i],f[j]+1);
}
}
maxNum=(maxNum>f[i]?maxNum:f[i]);
}
return maxNum;
}