Find a Fixed Point in a Given Array _ GeeksforGeeks

5/12/12 Find a Fixed Point in a given array | GeeksforGeeks

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Find a Fixed Point in a given array

December 1, 2011

Given an array of n distinct integers sorted in ascending order, write a function that returns aFixed Point in the array, if there is any Fixed Point present in array, else returns -1. FixedPoint in an array is an index i such that arr[i] is equal to i. Note that integers in array can benegative.

Examples:

Input: arr[] = {-10, -5, 0, 3, 7} Output: 3 // arr[3] == 3

Input: arr[] = {0, 2, 5, 8, 17} Output: 0 // arr[0] == 0



Input: arr[] = {-10, -5, 3, 4, 7, 9} Output: -1 // No Fixed Point

Asked by rajk

Method 1 (Linear Search) Linearly search for an index i such that arr[i] == i. Return the first such index found. Thanksto pm for suggesting this solution.

Time Complexity: O(n)

Method 2 (Binary Search)First check whether middle element is Fixed Point or not. If it is, then return it; otherwisecheck whether index of middle element is greater than value at the index. If index is greater,then Fixed Point(s) lies on the right side of the middle point (obviously only if there is a FixedPoint). Else the Fixed Point(s) lies on left side.

int linearSearch(int arr[], int n){ int i; for(i = 0; i < n; i++) { if(arr[i] == i) return i; } /* If no fixed point present then return -1 */ return -1;} /* Driver program to check above functions */int main(){ int arr[10] = {-10, -1, 0, 3, 10, 11, 30, 50, 100}; int n = sizeof(arr)/sizeof(arr[0]); printf("Fixed Point is %d", linearSearch(arr, n)); getchar(); return 0;}

int binarySearch(int arr[], int low, int high){ if(high >= low) { int mid = (low + high)/2; /*low + (high - low)/2;*/



Algorithmic Paradigm: Divide & ConquerTime Complexity: O(Logn)

Please write comments if you find anything incorrect, or you want to share more informationabout the topic discussed above.

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25 comments so far

1. Anon says:

if(mid == arr[mid]) return mid; if(mid > arr[mid]) return binarySearch(arr, (mid + 1), high); else return binarySearch(arr, low, (mid -1)); } /* Return -1 if there is no Fixed Point */ return -1;} /* Driver program to check above functions */int main(){ int arr[10] = {-10, -1, 0, 3, 10, 11, 30, 50, 100}; int n = sizeof(arr)/sizeof(arr[0]); printf("Fixed Point is %d", binarySearch(arr, 0, n-1)); getchar(); return 0;}

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April 21, 2012 at 6:40 PM

The binary search method will not work in case of duplicate elements in the array,because we can't eliminate one part of the array in each iteration.

Reply2. Avi says:

April 12, 2012 at 8:17 PM

Reply3. PsychoCoder says:

March 10, 2012 at 1:14 PM

In the function this portion is fixed irrespective of the given data.

So, if my input is

#include<stdio.h>#include<conio.h>int main(){ int arr[] = {-10, -5, 3, 4, 7, 9}; int len = sizeof(arr)/sizeof(int); static int flag; int num = -1; for(int i = 0; i<len; i++) { if(arr[i] == i) { flag = 1; num = arr[i]; break; } } printf("%d \n", num); getch(); }

if(mid > arr[mid]) return binarySearch(arr, (mid + 1), high);else return binarySearch(arr, low, (mid -1));



then 5(mid) < 11 (arr[mid]) so it will neglect right hand side the value '9' which isplaced at arr[9].

ReplyYogesh Batra says:April 8, 2012 at 4:12 PM

This is clearly given that sequence is in increasing order, find some other counterexample for binary look up method.

Reply4. kartikaditya says:

February 20, 2012 at 11:22 AM

#include <iostream>#include <stdio.h>

using namespace std;

int getFixedPoint(int a[], int n) {int start = 0, end = n - 1, mid = 0;while (start <= end) {mid = (start + end) >> 1;if (a[mid] == mid) {return mid;} else if (a[mid] < mid) {start = mid + 1;} else {end = mid - 1;}}return -1;}

int main() {int a1[] = {-10, -5, 0, 3, 7};cout << getFixedPoint(a1, 5) << endl;int a2[] = {0, 2, 5, 8, 17};cout << getFixedPoint(a2, 5) << endl;int a3[] = {-10, -5, 3, 4, 7, 9};cout << getFixedPoint(a3, 6) << endl;int a4[] = {-10, 1, 2, 3, 10, 11, 30, 50, 100};cout << getFixedPoint(a4, 9) << endl;

int arr[10] = {-10, -1, 0, 6, 10, 11, 30, 50, 100, 9};



return 0;}

Reply5. vijay_kansal says:

February 15, 2012 at 3:25 PM

An optimized algorithm for the case when elements can be repeated in the array :

The algo will return 1st element that is a fixed pt.

Replyvijay_kansal says:February 19, 2012 at 10:35 AM

An error correction,Replace the if condition of while loop by :

Reply6. Nages says:

February 10, 2012 at 6:55 AM

int fixed_pt(int *a,int n){ int l=0,h=n-1,m,i; while(l<=h) { m=(l+h)/2; if(a[m]>=0) h=m-1; else l=m+1; }// finding the 1st element(it will be a[l]) >=0 for(i=l;i<n;i=a[i]) { if(a[i]==i) return i; } return -1;}

if(a[m]>=m)

public class FixedPointInArray { public static void main(String[] arg) {



Reply7. Karthick says:

December 27, 2011 at 6:19 PM

I think, the following code can find the first Fixed Point.It would be good if someone can check this code for correctness.

int findFirstFixedPoint(int a[],int length){if(length==0){ return -1; }int start=0,end=length-1,middle;while(start<=end){middle=(start+end)/2;if(a[middle]==middle){

int arr[] = new int[] { -10, -5, 0, 3, 7 }; System.out.println(fixedPoint(arr)); arr = new int[] { 0, 2, 5, 8, 17 }; System.out.println(fixedPoint(arr)); arr = new int[] { -10, -5, 3, 4, 7, 9 }; System.out.println(fixedPoint(arr)); } public static int fixedPoint(int[] arr) { int pos = -1; for (int i = 0; i < arr.length;) { if (i == arr[i]) return i; else if (i < arr[i]) i = arr[i]; else ++i; } return pos; }}



if(middle==start || a[middle-1]!=middle-1){ return middle; }end=middle-1;}else if(a[middle]<middle){ start=middle+1; }else{ end=middle-1; }}return -1;}

ReplyKarthick says:December 27, 2011 at 6:25 PM

Sorry for posting without the sourcecode tags

Reply8. kaka09 says:


int findFirstFixedPoint(int length){ if(length==0){ return -1; } int start=0,end=length-1,middle; while(start<=end) { middle=(start+end)/2; if(a[middle]==middle) { if(middle==start || a[middle-1]!=middle-1){ return end=middle-1; } else if(a[middle]<middle){ start=middle+1; } else{ end=middle-1; } } return -1;}

#include<stdio.h>int main{ int n,i,j,flag=-1,pos; scanf("%d",&n);() int a[n]; for(i=0;i<n;i++) scanf("%d",&a[i]); i=0;



Reply9. kaka09 says:


Reply10. punit says:

while(a[i++]<0) for(j=i;j<n;j++) { if(j==a[j]) { flag=0; pos=j; break; } } if(flag!=0) printf("number not found!!"); else printf("number found at index %d\n", pos);}

#include<stdio.h>int main(){ int n,i,j,flag=-1,pos; scanf("%d",&n); int a[n]; for(i=0;i<n;i++) scanf("%d",&a[i]); i=0; while(a[i++]<0) for(j=i;j<n;j++) { if(j==a[j]) { flag=0; pos=j; break; } } if(flag!=0) printf("number not found!!"); else printf("number found at index %d\n", pos);}




for handling equal values like [-1, 0, 4, 4, 4, 4, 4] just tweak binary search as follows-int bs(vector v, int l, int r){int m = (l+r)/2;if(l > r)return -1;if(v[m] == m)return m;if((v[m] > m) && (v[m] != v.at(m+1))){return bs(v, l, m-1);}elsereturn bs(v, m+1, r);}

ReplyAG says:December 29, 2011 at 2:56 PM

I think it fails for this test case:-

[0, 2, 5, 5, 5, 6, 7]

ReplyYogesh Batra says:April 8, 2012 at 4:22 PM

It's working, I don't think there is any problem with the above code.

Replyyogendra says:April 9, 2012 at 9:31 PM

@YogeshI checked the above code is failing in case of [0, 2, 5, 5, 6, 7]in first Itrl=0,r=5 => m=2;so a[2]>2 and here a[2]==a[3]==5so l=3;

/* Paste your code here (You may delete these lines if not writing code) */



in second Itr l=3,r=5 =>m=4;s0 a[4]>6 hence it will return -1; int next iter.

Reply11. punit says:


/* Paste your code here (You may delete these lines if not writing code) */[#includeusing namespace std;#includeint bs(vector v, int l, int r){int m = (l+r)/2;if(l > r)return -1;if(v[m] == m)return m;if(v[m] > m){return bs(v, l, m-1);}elsereturn bs(v, m+1, r);}void Fp(vector v, int l, int r){int index;static int flag = true, Index = 0;index = r;while(1){index = bs(v, l, index-1);

if(index == -1){if(flag == false){cout<<"not found.\n";return;}cout<<"found at index = "<<Index<<endl;break;}



Index = index;flag = true;}}int main(){vector v;v.push_back(-1);v.push_back(0);v.push_back(2);v.push_back(5);Fp(v, 0, v.size());return 0;}]

Reply12. Daniel says:


The non-distinct case would be interesting.

Replykartik says:December 1, 2011 at 8:49 PM

I don't think binary search can be applied in case of non-distict case. Forexample, {-1, 4, 4, 4, 4, 4, 4}. We go to middle index 3 and see value is greaterthen index. We can't have a logic to look for left side or right side as array couldbe {-1, 0, 2, 4, 6, 8, 10} or {-1, 4, 4, 4, 5, 5, 5}. Let me know your thoughts.

Reply13. krishna says:


The problem would be more interesting if we ask for the first Fixed Point instead of aFixed Point.

Replykartik says:December 1, 2011 at 12:20 PM

@Krishna: Thanks for suggesting a more advanced version of the problem. Themethod 1 anyways finds the first FP. The method 2 can also be tweaked to findthe first FP.



ReplyVenki says:December 1, 2011 at 4:22 PM

@krishna, Good question. Binary search is such a powerful technique, sometimes, we miss to note its power. Solution to your question is given below. Letme know if there are any corner cases.

ReplyJing says:December 2, 2011 at 11:11 AM

For A=[-1, 0, 2], it seems to return -1 instead of 2, doesn't it?

ReplyVenki says:December 1, 2011 at 4:23 PM

Driver code.

int binarySearch(int A[], int l, int r){ int mid; int traced = -1; while( (r - l) > 1 ) { mid = l + (r - l)/2; if( A[mid] == mid ) traced = mid; // C/C++ smart enough to confuse (mid <= A[mid] ? r : l) = mid; } if( A[l] == l )// Corner case : ) traced = l; return traced;}

/* Paste your code here (You may delete these lines if not writing code) */

int main(){



ReplyVenki says:December 2, 2011 at 11:41 AM

@Jing, thanks for pointing the error. Given below is updated one.

int arr[] = {-10, 1, 2, 3, 10, 11, 30, 50, 100}; int n = sizeof(arr)/sizeof(arr[0]); printf("Fixed Point is %d", binarySearch(arr, 0, n-1)); return 0;}

#include <stdio.h> int binarySearchLeft(int A[], int l, int r){ int mid; int traced = -1; while( (r - l) > 1 ) { mid = l + (r - l)/2; if( A[mid] == mid ) traced = mid; (mid <= A[mid] ? r : l) = mid; } if( A[r] == r ) traced = r; if( A[l] == l ) traced = l; return traced;} int binarySearchRight(int A[], int l, int r){ int mid; int traced = -1; while( (r - l) > 1 ) { mid = l + (r - l)/2; if( A[mid] == mid ) traced = mid;



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(mid >= A[mid] ? l : r) = mid; } if( A[l] == l ) traced = l; if( A[r] == r ) traced = r; return traced;} int main(){ int arr[] = { 0, 1, 2 }; int n = sizeof(arr)/sizeof(arr[0]); printf("Fixed Point is %d\n", binarySearchLeft(arr, 0, n-1)); printf("Fixed Point is %d\n", binarySearchRight(arr, 0, n-1)); return 0;}



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