The name of this searching algorithm may be misleading as it works in O(Log n) time. The name comes from the way it searches an element.
Given a sorted array, and an element x to be searched, find position of x in the array.
Input: arr[] = {10, 20, 40, 45, 55}
x = 45
Output: Element found at index 3
Input: arr[] = {10, 15, 25, 45, 55}
x = 15
Output: Element found at index 1
We have discussed, linear search, binary search for this problem.
Exponential search involves two steps:
- Find range of indexes where element is present
- Do Binary Search in above found range.
How to find the range where element may be present?
The idea is to start with subarray size 1, compare its last element with x, then try size 2, then 4 and so on until last element of a subarray is not greater.
Once we find an index i (after repeated doubling of i), we know that the element must be present between i/2 and i (Why i/2? because we could not find a greater value in previous iteration)
Recursive Implementation - O(Log n) Time and O(Log n) Space
We start with an index i equal to 1 and repeatedly double it until either i is greater than or equal to the length of the array or the value at index i is greater than or equal to the target value x. We then perform a binary search on the range [i/2, min(i, n-1)], where n is the length of the array. This range is guaranteed to contain the target value, if it is present in the array, because we know that the target value must be greater than or equal to the value at index i/2 and less than or equal to the value at index min(i, n-1). If we find the target value in the binary search, we return its index. Otherwise, we return -1 to indicate that the target value is not present in the array.
We mainly use recursive implementation of binary search once we find the range. We use iterative code to find the range.
C++
#include <bits/stdc++.h>
using namespace std;
int binarySearch(vector<int>& arr, int l, int r, int x);
// Returns position of first occurrence of
// x in array
int exponentialSearch(vector<int>& arr, int n, int x)
{
// If x is present at first location itself
if (arr[0] == x)
return 0;
// Find range for binary search by
// repeated doubling
int i = 1;
while (i < n && arr[i] <= x)
i = i * 2;
// Call binary search for the found range.
return binarySearch(arr, i / 2, min(i, n - 1), x);
}
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is
// present, otherwise -1
int binarySearch(vector<int>& arr, int l, int r, int x)
{
if (r >= l)
{
int mid = l + (r - l) / 2;
// If the element is present at the middle
// itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then it
// can only be present n left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only be present
// in right subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element is not present
// in array
return -1;
}
// Driver code
int main(void)
{
vector<int> arr = {2, 3, 4, 10, 40};
int n = arr.size();
int x = 10;
int result = exponentialSearch(arr, n, x);
(result == -1) ? cout << "Element is not present in array"
: cout << "Element is present at index " << result;
return 0;
}
// C++ program to find an element x in a
// sorted array using Exponential search.
#include <stdio.h>
#include <time.h>
#include <math.h>
#define min
int binarySearch(int arr[], int, int, int);
// Returns position of first occurrence of
// x in array
int exponentialSearch(int arr[], int n, int x)
{
// If x is present at first location itself
if (arr[0] == x)
return 0;
// Find range for binary search by
// repeated doubling
int i = 1;
while (i < n && arr[i] <= x)
i = i*2;
// Call binary search for the found range.
return binarySearch(arr, i/2,
min(i, n-1), x);
}
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is
// present, otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l)
{
int mid = l + (r - l)/2;
// If the element is present at the middle
// itself
if (arr[mid] == x)
return mid;
// If element is smaller than mid, then it
// can only be present n left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid-1, x);
// Else the element can only be present
// in right subarray
return binarySearch(arr, mid+1, r, x);
}
// We reach here when element is not present
// in array
return -1;
}
// Driver code
int main(void)
{
int arr[] = {2, 3, 4, 10, 40};
int n = sizeof(arr)/ sizeof(arr[0]);
int x = 10;
int result = exponentialSearch(arr, n, x);
(result == -1)? printf("Element is not
present in array")
: printf("Element is present
at index %d",
result);
return 0;
}
// Java program to
// find an element x in a
// sorted array using
// Exponential search.
import java.util.Arrays;
class GFG
{
// Returns position of
// first occurrence of
// x in array
static int exponentialSearch(int arr[],
int n, int x)
{
// If x is present at first location itself
if (arr[0] == x)
return 0;
// Find range for binary search by
// repeated doubling
int i = 1;
while (i < n && arr[i] <= x)
i = i*2;
// Call binary search for the found range.
return Arrays.binarySearch(arr, i/2,
Math.min(i, n-1), x);
}
// Driver code
public static void main(String args[])
{
int arr[] = {2, 3, 4, 10, 40};
int x = 10;
int result = exponentialSearch(arr,
arr.length, x);
System.out.println((result < 0) ?
"Element is not present in array" :
"Element is present at index " +
result);
}
}
# Python program to find an element x
# in a sorted array using Exponential Search
# A recursive binary search function returns
# location of x in given array arr[l..r] is
# present, otherwise -1
def binarySearch( arr, l, r, x):
if r >= l:
mid = l + ( r-l ) // 2
# If the element is present at
# the middle itself
if arr[mid] == x:
return mid
# If the element is smaller than mid,
# then it can only be present in the
# left subarray
if arr[mid] > x:
return binarySearch(arr, l,
mid - 1, x)
# Else he element can only be
# present in the right
return binarySearch(arr, mid + 1, r, x)
# We reach here if the element is not present
return -1
# Returns the position of first
# occurrence of x in array
def exponentialSearch(arr, n, x):
# IF x is present at first
# location itself
if arr[0] == x:
return 0
# Find range for binary search
# j by repeated doubling
i = 1
while i < n and arr[i] <= x:
i = i * 2
# Call binary search for the found range
return binarySearch( arr, i // 2,
min(i, n-1), x)
# Driver Code
arr = [2, 3, 4, 10, 40]
n = len(arr)
x = 10
result = exponentialSearch(arr, n, x)
if result == -1:
print ("Element not found in the array")
else:
print ("Element is present at index %d" %(result))
# This code is contributed by Harshit Agrawal
// C# program to find an element x in a
// sorted array using Exponential search.
using System;
class GFG {
// Returns position of first
// occurrence of x in array
static int exponentialSearch(int []arr,
int n, int x)
{
// If x is present at
// first location itself
if (arr[0] == x)
return 0;
// Find range for binary search
// by repeated doubling
int i = 1;
while (i < n && arr[i] <= x)
i = i * 2;
// Call binary search for
// the found range.
return binarySearch(arr, i/2,
Math.Min(i, n - 1), x);
}
// A recursive binary search
// function. It returns location
// of x in given array arr[l..r] is
// present, otherwise -1
static int binarySearch(int []arr, int l,
int r, int x)
{
if (r >= l)
{
int mid = l + (r - l) / 2;
// If the element is present
// at the middle itself
if (arr[mid] == x)
return mid;
// If element is smaller than
// mid, then it can only be
// present n left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only
// be present in right subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element
// is not present in array
return -1;
}
// Driver code
public static void Main()
{
int []arr = {2, 3, 4, 10, 40};
int n = arr.Length;
int x = 10;
int result = exponentialSearch(arr, n, x);
if (result == -1)
Console.Write("Element is not
present in array");
else
Console.Write("Element is
present at index "
+ result);
}
}
// This code is contributed by Smitha
<script>
// Javascript program to find an element x
// in a sorted array using Exponential Search
// A recursive binary search
// function. It returns location
// of x in given array arr[l..r] is
// present, otherwise -1
function binarySearch(arr, l, r, x)
{
if (r >= l)
{
let mid = l + (r - l) / 2;
// If the element is present
// at the middle itself
if (arr[mid] == x)
return mid;
// If element is smaller than
// mid, then it can only be
// present n left subarray
if (arr[mid] > x)
return binarySearch(arr, l, mid - 1, x);
// Else the element can only
// be present in right subarray
return binarySearch(arr, mid + 1, r, x);
}
// We reach here when element
// is not present in array
return -1;
}
// Returns position of first
// occurrence of x in array
function exponentialSearch(arr, n, x)
{
// If x is present at
// first location itself
if (arr[0] == x)
return 0;
// Find range for binary search
// by repeated doubling
let i = 1;
while (i < n && arr[i] <= x)
i = i * 2;
// Call binary search for
// the found range.
return binarySearch(arr, i/2,
Math.min(i, n - 1), x);
}
// Driver Code
let arr = [2, 3, 4, 10, 40];
let n = arr.length;
let x = 10;
let result = exponentialSearch(arr, n, x);
if (result == -1)
document.write("Element is not present in array");
else
document.write("Element is present at index " + result);
</script>
<?php
// PHP program to find an element x in a
// sorted array using Exponential search.
// Returns position of first
// occurrence of x in array
function exponentialSearch($arr, $n, $x)
{
// If x is present at
// first location itself
if ($arr[0] == $x)
return 0;
// Find range for binary search
// by repeated doubling
$i = 1;
while ($i< $n and $arr[$i] <=$x)
$i = $i * 2;
// Call binary search
// for the found range.
return binarySearch($arr, $i / 2,
min($i, $n - 1), $x);
}
// A recursive binary search
// function. It returns location
// of x in given array arr[l..r] is
// present, otherwise -1
function binarySearch($arr, $l,
$r, $x)
{
if ($r >= $l)
{
$mid = $l + ($r - $l)/2;
// If the element is
// present at the middle
// itself
if ($arr[$mid] == $x)
return $mid;
// If element is smaller
// than mid, then it
// can only be present
// n left subarray
if ($arr[$mid] > $x)
return binarySearch($arr, $l,
$mid - 1, $x);
// Else the element
// can only be present
// in right subarray
return binarySearch($arr, $mid + 1,
$r, $x);
}
// We reach here when
// element is not present
// in array
return -1;
}
// Driver code
$arr = array(2, 3, 4, 10, 40);
$n = count($arr);
$x = 10;
$result = exponentialSearch($arr, $n, $x);
if($result == -1)
echo "Element is not present in array";
else
echo "Element is present at index ",
$result;
// This code is contributed by anuj_67
?>
OutputElement is present at index 3
Iterative Implementation - O(Log n) Time and O(1) Space
Here we use iterative implementation of binary search with the same approach.
C++
#include<bits/stdc++.h>
using namespace std;
int exponential_search(vector<int> arr,int x){
int n = arr.size();
if(n == 0)
return -1;
// Find range for binary search by repeatedly doubling i
int i = 1;
while(i < n and arr[i] < x)
i *= 2;
// Perform binary search on the range [i/2, min(i, n-1)]
int left = i/2;
int right = min(i, n-1);
while(left <= right){
int mid = (left + right)/2;
if(arr[mid] == x) return mid;
else if(arr[mid] < x) left = mid + 1;
else right = mid - 1;
}
return -1;
}
// Driver Code
int main(){
vector<int> arr = {2, 3, 4, 10, 40};
int n = arr.size();
int x = 10;
int result = exponential_search(arr, x);
if(result == -1){
cout << "Element not found in the array";
}
else{
cout << "Element is present at index " << result << endl;
}
return 0;
}
// This code is contributed by Ajay singh
// Java implementation of above approach
import java.util.*;
class Main {
// Exponential search function
public static int exponential_search(ArrayList<Integer> arr, int x) {
int n = arr.size();
if (n == 0)
return -1;
// Find range for binary search by repeatedly doubling i
int i = 1;
while (i < n && arr.get(i) < x)
i *= 2;
// Perform binary search on the range [i/2, min(i, n-1)]
int left = i / 2;
int right = Math.min(i, n - 1);
while (left <= right) {
int mid = (left + right) / 2; // finding mid
if (arr.get(mid) == x)
return mid;
else if (arr.get(mid) < x)
left = mid + 1;
else
right = mid - 1;
}
return -1;
}
// Driver Code
public static void main(String[] args) {
ArrayList<Integer> arr = new ArrayList<>(Arrays.asList(2, 3, 4, 10, 40));
int n = arr.size();
int x = 10;
int result = exponential_search(arr, x);
if (result == -1) {
System.out.println("Element not found in the array");
}
else {
System.out.println("Element is present at index " + result);
}
}
}
def exponential_search(arr, x):
n = len(arr)
if n == 0:
return -1
# Find range for binary search by repeatedly doubling i
i = 1
while i < n and arr[i] < x:
i *= 2
# Perform binary search on the range [i/2, min(i, n-1)]
left = i // 2
right = min(i, n-1)
while left <= right:
mid = (left + right) // 2
if arr[mid] == x:
return mid
elif arr[mid] < x:
left = mid + 1
else:
right = mid - 1
return -1
# Driver Code
arr = [2, 3, 4, 10, 40]
n = len(arr)
x = 10
result = exponential_search(arr, x)
if result == -1:
print ("Element not found in the array")
else:
print ("Element is present at index %d" %(result))
# This code is contributed by Ajay singh
// C# Program for the above approach
using System;
using System.Collections.Generic;
class Program
{
static int exponential_search(List<int> arr, int x)
{
int n = arr.Count;
if (n == 0)
return -1;
// Find range for binary search by repeatedly doubling i
int i = 1;
while (i < n && arr[i] < x)
i *= 2;
// Perform binary search on the range [i/2, min(i, n-1)]
int left = i / 2;
int right = Math.Min(i, n - 1);
while (left <= right)
{
int mid = (left + right) / 2;
if (arr[mid] == x) return mid;
else if (arr[mid] < x) left = mid + 1;
else right = mid - 1;
}
return -1;
}
static void Main(string[] args)
{
List<int> arr = new List<int> { 2, 3, 4, 10, 40 };
int n = arr.Count;
int x = 10;
int result = exponential_search(arr, x);
if (result == -1)
{
Console.WriteLine("Element not found in the array");
}
else
{
Console.WriteLine("Element is present at index " + result);
}
}
}
// This code is contributed by princekumaras
function exponential_search(arr, x) {
const n = arr.length;
if (n == 0) { // if array is empty, return -1
return -1;
}
let i = 1;
while (i < n && arr[i] < x) { // Find the range for binary search by repeatedly doubling i
i *= 2;
}
const left = Math.floor(i / 2); // Set left boundary for binary search
const right = Math.min(i, n - 1); // Set right boundary for binary search
while (left <= right) { // Perform binary search on the range [i/2, min(i, n-1)]
const mid = Math.floor((left + right) / 2); // Find middle index
if (arr[mid] == x) { // If element is found at mid index, return mid
return mid;
} else if (arr[mid] < x) { // If element is less than mid index, search the right half of array
left = mid + 1;
} else { // If element is greater than mid index, search the left half of array
right = mid - 1;
}
}
return -1; // If element is not found in array, return -1
}
// Driver Code
const arr = [2, 3, 4, 10, 40];
const n = arr.length;
const x = 10;
const result = exponential_search(arr, x);
if (result == -1) {
console.log("Element not found in the array");
} else {
console.log(`Element is present at index ${result}`);
}
OutputElement is present at index 3
Applications of Exponential Search:
- Exponential Binary Search is particularly useful for unbounded searches, where size of array is infinite. Please refer Unbounded Binary Search for an example.
- It works better than Binary Search for bounded arrays when the element to be searched is closer to the first element.
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Bitonic Point - Maximum in Increasing Decreasing ArrayGiven an array arr[] of integers which is initially strictly increasing and then strictly decreasing, the task is to find the bitonic point, that is the maximum value in the array. Note: Bitonic Point is a point in bitonic sequence before which elements are strictly increasing and after which elemen
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Given Array of size n and a number k, find all elements that appear more than n/k timesGiven an array of size n and an integer k, find all elements in the array that appear more than n/k times. Examples:Input: arr[ ] = [3, 4, 2, 2, 1, 2, 3, 3], k = 4Output: [2, 3]Explanation: Here n/k is 8/4 = 2, therefore 2 appears 3 times in the array that is greater than 2 and 3 appears 3 times in
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Medium problems on Searching algorithms
3 Sum - Find All Triplets with Zero SumGiven an array arr[], the task is to find all possible indices {i, j, k} of triplet {arr[i], arr[j], arr[k]} such that their sum is equal to zero and all indices in a triplet should be distinct (i != j, j != k, k != i). We need to return indices of a triplet in sorted order, i.e., i < j < k.Ex
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Find the element before which all the elements are smaller than it, and after which all are greaterGiven an array, find an element before which all elements are equal or smaller than it, and after which all the elements are equal or greater.Note: Print -1, if no such element exists.Examples:Input: arr[] = [5, 1, 4, 3, 6, 8, 10, 7, 9]Output: 6 Explanation: 6 is present at index 4. All elements on
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Largest pair sum in an arrayGiven an unsorted of distinct integers, find the largest pair sum in it. For example, the largest pair sum is 74. If there are less than 2 elements, then we need to return -1.Input : arr[] = {12, 34, 10, 6, 40}, Output : 74Input : arr[] = {10, 10, 10}, Output : 20Input arr[] = {10}, Output : -1[Naiv
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K’th Smallest Element in Unsorted ArrayGiven an array arr[] of N distinct elements and a number K, where K is smaller than the size of the array. Find the K'th smallest element in the given array. Examples:Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 3 Output: 7Input: arr[] = {7, 10, 4, 3, 20, 15}, K = 4 Output: 10 Table of Content[Naive Ap
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Search in a Sorted and Rotated ArrayGiven a sorted and rotated array arr[] of n distinct elements, the task is to find the index of given key in the array. If the key is not present in the array, return -1. Examples: Input: arr[] = [5, 6, 7, 8, 9, 10, 1, 2, 3], key = 3Output: 8Explanation: 3 is present at index 8 in arr[].Input: arr[]
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Minimum in a Sorted and Rotated ArrayGiven a sorted array of distinct elements arr[] of size n that is rotated at some unknown point, the task is to find the minimum element in it. Examples: Input: arr[] = [5, 6, 1, 2, 3, 4]Output: 1Explanation: 1 is the minimum element present in the array.Input: arr[] = [3, 1, 2]Output: 1Explanation:
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Find a Fixed Point (Value equal to index) in a given arrayGiven an array of n distinct integers sorted in ascending order, the task is to find the First Fixed Point in the array. Fixed Point in an array is an index i such that arr[i] equals i. Note that integers in the array can be negative. Note: If no Fixed Point is present in the array, print -1.Example
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K Mmost Frequent Words in a FileGiven a book of words and an integer K. Assume you have enough main memory to accommodate all words. Design a dynamic data structure to find the top K most frequent words in a book. The structure should allow new words to be added in main memory.Examples:Input: fileData = "Welcome to the world of Ge
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Closest K Elements in a Sorted ArrayYou are given a sorted array arr[] containing unique integers, a number k, and a target value x. Your goal is to return exactly k elements from the array that are closest to x, excluding x itself if it is present in the array.An element a is closer to x than b if:|a - x| < |b - x|, or|a - x| == |
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2 Sum - Pair Sum Closest to Target using Binary SearchGiven an array arr[] of n integers and an integer target, the task is to find a pair in arr[] such that it’s sum is closest to target.Note: Return the pair in sorted order and if there are multiple such pairs return the pair with maximum absolute difference. If no such pair exists return an empty ar
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Find the closest pair from two sorted arraysGiven two arrays arr1[0...m-1] and arr2[0..n-1], and a number x, the task is to find the pair arr1[i] + arr2[j] such that absolute value of (arr1[i] + arr2[j] - x) is minimum. Example: Input: arr1[] = {1, 4, 5, 7}; arr2[] = {10, 20, 30, 40}; x = 32Output: 1 and 30Input: arr1[] = {1, 4, 5, 7}; arr2[]
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Find three closest elements from given three sorted arraysGiven three sorted arrays A[], B[] and C[], find 3 elements i, j and k from A, B and C respectively such that max(abs(A[i] - B[j]), abs(B[j] - C[k]), abs(C[k] - A[i])) is minimized. Here abs() indicates absolute value. Example : Input : A[] = {1, 4, 10} B[] = {2, 15, 20} C[] = {10, 12} Output: 10 15
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Search in an Array of Rational Numbers without floating point arithmeticGiven a sorted array of rational numbers, where each rational number is represented in the form p/q (where p is the numerator and q is the denominator), the task is to find the index of a given rational number x in the array. If the number does not exist in the array, return -1.Examples: Input: arr[
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Hard problems on Searching algorithms
Median of two sorted arrays of same sizeGiven 2 sorted arrays a[] and b[], each of size n, the task is to find the median of the array obtained after merging a[] and b[]. Note: Since the size of the merged array will always be even, the median will be the average of the middle two numbers.Input: a[] = [1, 12, 15, 26, 38], b[] = [2, 13, 17
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Search in an almost sorted arrayGiven a sorted integer array arr[] consisting of distinct elements, where some elements of the array are moved to either of the adjacent positions, i.e. arr[i] may be present at arr[i-1] or arr[i+1].Given an integer target. You have to return the index ( 0-based ) of the target in the array. If targ
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Find position of an element in a sorted array of infinite numbersGiven a sorted array arr[] of infinite numbers. The task is to search for an element k in the array.Examples:Input: arr[] = [3, 5, 7, 9, 10, 90, 100, 130, 140, 160, 170], k = 10Output: 4Explanation: 10 is at index 4 in array.Input: arr[] = [2, 5, 7, 9], k = 3Output: -1Explanation: 3 is not present i
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Pair Sum in a Sorted and Rotated ArrayGiven an array arr[] of size n, which is sorted and then rotated around an unknown pivot, the task is to check whether there exists a pair of elements in the array whose sum is equal to a given target value.Examples : Input: arr[] = [11, 15, 6, 8, 9, 10], target = 16Output: trueExplanation: There is
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K’th Smallest/Largest Element in Unsorted Array | Worst case Linear TimeGiven an array of distinct integers arr[] and an integer k. The task is to find the k-th smallest element in the array. For better understanding, k refers to the element that would appear in the k-th position if the array were sorted in ascending order. Note: k will always be less than the size of t
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K'th largest element in a streamGiven an input stream of n integers, represented as an array arr[], and an integer k. After each insertion of an element into the stream, you need to determine the kth largest element so far (considering all elements including duplicates). If k elements have not yet been inserted, return -1 for that
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Best First Search (Informed Search)Best First Search is a heuristic search algorithm that selects the most promising node for expansion based on an evaluation function. It prioritizes nodes in the search space using a heuristic to estimate their potential. By iteratively choosing the most promising node, it aims to efficiently naviga
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