I am attempting to expand a function to find the number of integer matches through Binary Search by resetting the high variable, but it gets stuck in a loop. I am guessing a workaround would be to duplicate this function to obtain the last index to determine the number of matches, but I do not think this would be such an elegant solution.
From this:
public static Matches findMatches(int[] values, int query) {
int firstMatchIndex = -1;
int lastMatchIndex = -1;
int numberOfMatches = 0;
int low = 0;
int mid = 0;
int high = values[values.length - 1];
boolean searchFirst = false;
while (low <= high){
mid = (low + high)/2;
if (values[mid] == query && firstMatchIndex == -1){
firstMatchIndex = mid;
if (searchFirst){
high = mid - 1;
searchFirst = false;
} else {
low = mid + 1;
}
} else if (query < values[mid]){
high = mid - 1;
} else {
low = mid + 1;
}
}
if (firstMatchIndex != -1) { // First match index is set
return new Matches(firstMatchIndex, numberOfMatches);
}
else { // First match index is not set
return new Matches(-1, 0);
}
}
To something like this:
public static Matches findMatches(int[] values, int query) {
int firstMatchIndex = -1;
int lastMatchIndex = -1;
int numberOfMatches = 0;
int low = 0;
int mid = 0;
int high = values[values.length - 1];
boolean searchFirst = false;
while (low <= high){
mid = (low + high)/2;
if (values[mid] == query && firstMatchIndex == -1){
firstMatchIndex = mid;
if (searchFirst){
high = values[values.length - 1]; // This is stuck in a loop
searchFirst = false;
}
} else if (values[mid] == query && lastMatchIndex == -1){
lastMatchIndex = mid;
if (!searchFirst){
high = mid - 1;
} else {
low = mid + 1;
}
} else if (query < values[mid]){
high = mid - 1;
} else {
low = mid + 1;
}
}
if (firstMatchIndex != -1) { // First match index is set
return new Matches(firstMatchIndex, numberOfMatches);
}
else { // First match index is not set
return new Matches(-1, 0);
}
}
There is a problem with your code:
high = values[values.length - 1];
should be
high = values.length - 1;
Also you do not need variables like numberOfMatches and searchFirst, we can have rather simple solution.
Now coming to the problem,I understand what you want I think Binary Search is appropriate for such query.
The Best way to do the required is once a match is found you just go forward and backward from that index until a mismatch occurs and this would be both elegant and efficient in calculating the firstMatchIndex and numberOfMatches.
So your function should be:
public static Matches findMatches(int[] values, int query)
{
int firstMatchIndex = -1,lastMatchIndex=-1;
int low = 0,mid = 0,high = values.length - 1;
while (low <= high)
{
mid = (low + high)/2;
if(values[mid]==query)
{
lastMatchIndex=mid;
firstMatchIndex=mid;
while(lastMatchIndex+1<values.length&&values[lastMatchIndex+1]==query)
lastMatchIndex++;
while(firstMatchIndex-1>=0&&values[firstMatchIndex-1]==query)
firstMatchIndex--;
return new Matches(firstMatchIndex,lastMatchIndex-firstMatchIndex+1);
}
else if(values[mid]>query)
high=mid-1;
else low=mid+1;
}
return new Matches(-1,0);
}
Couldn't you just use something like a set to find duplicates?
Something like this:
package example;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
public class DuplicatesExample {
public static void main(String[] args) {
String[] strings = { "one", "two", "two", "three", "four", "five", "six", "six" };
List<String> dups = getDups(strings);
System.out.println("DUPLICATES:");
for(String str : dups) {
System.out.println("\t" + str);
}
}
private static List<String> getDups(String[] strings) {
ArrayList<String> rtn = new ArrayList<String>();
HashSet<String> set = new HashSet<>();
for (String str : strings) {
boolean added = set.add(str);
if (added == false ) {
rtn.add(str);
}
}
return rtn;
}
}
Output:
DUPLICATES:
two
six
I have split your problems into two parts - using binary search to find a number and counting the number of matches. The first part is resolved by the search function while the second part is resolved by the findMatches function:
public static Matches findMatches(int[] values, int query) {
int leftIndex = -1;
int rightIndex = -1;
int high = values.length - 1;
int matchedIndex = search(values, 0, high, query);
//if at least one match
if (matchedIndex != -1) {
//decrement upper bound of left array
int leftHigh = matchedIndex - 1;
//increment lower bound of right array
int rightLow = matchedIndex + 1;
//loop until no more duplicates in left array
while (true) {
int leftMatchedIndex = search(values, 0, leftHigh, query);
//if duplicate found
if (leftMatchedIndex != -1) {
leftIndex = leftMatchedIndex;
//decrement upper bound of left array
leftHigh = leftMatchedIndex - 1;
} else {
break;
}
}
//loop until no more duplicates in right array
while(true){
int rightMatchedIndex = search(values, rightLow, high, query);
//if duplicate found
if(rightMatchedIndex != -1){
rightIndex = rightMatchedIndex;
//increment lower bound of right array
rightLow = rightMatchedIndex + 1;
} else{
break;
}
}
return new Matches(matchedIndex, rightIndex - leftIndex + 1);
}
return new Matches(-1, 0);
}
private static int search(int[] values, int low, int high, int query) {
while (low <= high) {
int mid = (low + high) / 2;
if (values[mid] == query) {
return mid;
} else if (query < values[mid]) {
high = mid - 1;
} else {
low = mid + 1;
}
}
return -1;
}
I found a solution after correcting a mistake with resetting high variable that caused an infinite loop.
public static Matches findMatches(int[] values, int query) {
int firstMatchIndex = -1;
int lastMatchIndex = -1;
int numberOfMatches = 0;
int low = 0;
int mid = 0;
int high = values.length - 1;
while (low <= high){
mid = (low + high)/2;
if (values[mid] == query && firstMatchIndex == -1){
firstMatchIndex = mid;
numberOfMatches++;
high = values.length - 1;
low = mid;
} else if (values[mid] == query && (lastMatchIndex == -1 || lastMatchIndex != -1)){
lastMatchIndex = mid;
numberOfMatches++;
if (query < values[mid]){
high = mid - 1;
} else {
low = mid + 1;
}
} else if (query < values[mid]){
high = mid - 1;
} else {
low = mid + 1;
}
}
if (firstMatchIndex != -1) { // First match index is set
return new Matches(firstMatchIndex, numberOfMatches);
}
else { // First match index is not set
return new Matches(-1, 0);
}
}
Not having any knowledge of the data other than sorted a priori is tough. See this: Binary Search O(log n) algorithm to find duplicate in sequential list?
This will find the first index of duplicates of k in a sorted array. Of course this is related to knowing the value of duplicate first but very useful when it is known.
public static int searchFirstIndexOfK(int[] A, int k) {
int left = 0, right = A.length - 1, result = -1;
// [left : right] is the candidate set.
while (left <= right) {
int mid = left + ((right - left) >>> 1); // left + right >>> 1;
if (A[mid] > k) {
right = mid - 1;
} else if (A[mid] == k) {
result = mid;
right = mid - 1; // Nothing to the right of mid can be
// solution.
} else { // A[mid] < k
left = mid + 1;
}
}
return result;
}
This will find a dupe in log(n) time but is fragile in that the data must be sorted as well as increasing by 1 and in the range 1..n.
static int findeDupe(int[] array) {
int low = 0;
int high = array.length - 1;
while (low <= high) {
int mid = (low + high) >>> 1;
if (array[mid] == mid) {
low = mid + 1;
} else {
high = mid - 1;
}
}
System.out.println("returning" + high);
return high;
}
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