I've below code snippet, A is non-decreasing order and need to check if X is present in A. If X is present in A then return position of occurrence of X in A otherwise returns -1. This code is not correct for some inputs, and need to fix it. Utmost I can modify 3 lines.
class Checking{
int check(int[] A, int X) {
int N = A.length;
if (N == 0) {
return -1;
}
int l = 0;
int r = N - 1;
while (l < r) {
int m = (l + r) / 2;
if (A[m] > X) {
r = m - 1;
} else {
l = m;
}
}
if (A[l] == X) {
return l;
}
return -1;
}
}
I couldnt figure out the fix, any suggestions will be helpful ?
Wiki page says that for such implementation of binary search middle element should be calculated using ceiling, so you can change to
int m = (l + r + 1) / 2;
Related
I'm always confused with Binary search right boarder.
For example, if we want to find the last index of target in a sorted array, the code should be:
public int binarySearchLarger(int[] nums, int target) {
int l = 0;
int r = nums.length - 1;
while (l < r) {
int m = l + (r - l + 1) / 2;
if (nums[m] < target) l = m;
else if(nums[m] > target) r = m - 1;
else l = m;
}
if (nums[l] == target) return l;
else return -1;
}
but I see lots of posts said that if we want while(l<r), then r should be nums.length, in this case, the code becomes:
public int binarySearchLarger(int[] nums, int target) {
int l = 0;
int r = nums.length;
while (l < r) {
int m = l + (r - l + 1) / 2;
if (nums[m] < target) l = m;
else if(nums[m] > target) r = m - 1;
else l = m;
}
if (nums[l] == target) return l;
else return -1;
}
but in this case, if (nums[m] < target) l = m; will throw arrayindexoutofbound exception.
My questions is: when should we use r = nums.length - 1 and when should we use r = nums.length?
See this answer for a discussion about the various options when implementing binary search: Binary Search algorithm implementations
For your problem, which seems to be to find the last instance of the target, I do it like this:
int findLastIndexOfTarget(int[] nums, int toFind) {
// find the position that divides <= toFind from > toFind
// these are the lowest and highest possible indexes
int low = 0;
int high = nums.length;
// while there is a range of possible indexes
while(low<high) {
int test = low+((high-low)/2);
if (nums[test] <= toFind) {
//too low
low = test+1; //guaranteed > low, <= high
} else {
//not too low
high = test; //guaranteed >= low, < high
}
}
// guaranteed low == high
// We found the position we were looking for. See if the target is on the left
return (low > 0 && nums[low-1]==target) ? low-1 : -1;
}
I do it like this:
int findEndIndex( vector<int> &nums, int target ){
int l = 0, r = nums.size() - 1;
while( l < r ){
int m = (l + (r - l) / 2) + 1;
if( nums[m] == target )
l = m;
else if( nums[m] < target )
l = m + 1;
else
r = m - 1;
}
return l >= nums.size() || nums[l] != target ? -1 : l;
}
Trying to solve this problem with recursion and memoization but for input 7168 I'm getting wrong answer.
public int numSquares(int n) {
Map<Integer, Integer> memo = new HashMap();
List<Integer> list = fillSquares(n, memo);
if (list == null)
return 1;
return helper(list.size()-1, list, n, memo);
}
private int helper(int index, List<Integer> list, int left, Map<Integer, Integer> memo) {
if (left == 0)
return 0;
if (left < 0 || index < 0)
return Integer.MAX_VALUE-1;
if (memo.containsKey(left)) {
return memo.get(left);
}
int d1 = 1+helper(index, list, left-list.get(index), memo);
int d2 = 1+helper(index-1, list, left-list.get(index), memo);
int d3 = helper(index-1, list, left, memo);
int d = Math.min(Math.min(d1,d2), d3);
memo.put(left, d);
return d;
}
private List<Integer> fillSquares(int n, Map<Integer, Integer> memo) {
int curr = 1;
List<Integer> list = new ArrayList();
int d = (int)Math.pow(curr, 2);
while (d < n) {
list.add(d);
memo.put(d, 1);
curr++;
d = (int)Math.pow(curr, 2);
}
if (d == n)
return null;
return list;
}
I'm calling like this:
numSquares(7168)
All test cases pass (even complex cases), but this one fails. I suspect something is wrong with my memoization but cannot pinpoint what exactly. Any help will be appreciated.
You have the memoization keyed by the value to be attained, but this does not take into account the value of index, which actually puts restrictions on which powers you can use to attain that value. That means that if (in the extreme case) index is 0, you can only reduce what is left with one square (1²), which rarely is the optimal way to form that number. So in a first instance memo.set() will register a non-optimal number of squares, which later will get updated by other recursive calls which are pending in the recursion tree.
If you add some conditional debugging code, you'll see that map.set is called for the same value of left multiple times, and with differing values. This is not good, because that means the if (memo.has(left)) block will execute for cases where that value is not guaranteed to be optimal (yet).
You could solve this by incorporating the index in your memoization key. This increases the space used for memoization, but it will work. I assume you can work this out.
But according to Lagrange's four square theorem every natural number can be written as the sum of at most four squares, so the returned value should never be 5 or more. You can shortcut the recursion when you get passed that number of terms. This reduces the benefit of using memoization.
Finally, there is a mistake in fillSquares: it should add n itself also when it is a perfect square, otherwise you'll not find solutions that should return 1.
Not sure about your bug, here is a short dynamic programming Solution:
Java
public class Solution {
public static final int numSquares(
final int n
) {
int[] dp = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
dp[0] = 0;
for (int i = 1; i <= n; i++) {
int j = 1;
int min = Integer.MAX_VALUE;
while (i - j * j >= 0) {
min = Math.min(min, dp[i - j * j] + 1);
++j;
}
dp[i] = min;
}
return dp[n];
}
}
C++
// Most of headers are already included;
// Can be removed;
#include <iostream>
#include <cstdint>
#include <vector>
#include <algorithm>
// The following block might slightly improve the execution time;
// Can be removed;
static const auto __optimize__ = []() {
std::ios::sync_with_stdio(false);
std::cin.tie(nullptr);
std::cout.tie(nullptr);
return 0;
}();
#define MAX INT_MAX
using ValueType = std::uint_fast32_t;
struct Solution {
static const int numSquares(
const int n
) {
if (n < 1) {
return 0;
}
static std::vector<ValueType> count_perfect_squares{0};
while (std::size(count_perfect_squares) <= n) {
const ValueType len = std::size(count_perfect_squares);
ValueType count_squares = MAX;
for (ValueType index = 1; index * index <= len; ++index) {
count_squares = std::min(count_squares, 1 + count_perfect_squares[len - index * index]);
}
count_perfect_squares.emplace_back(count_squares);
}
return count_perfect_squares[n];
}
};
int main() {
std::cout << std::to_string(Solution().numSquares(12) == 3) << "\n";
return 0;
}
Python
Here we can simply use lru_cache:
class Solution:
dp = [0]
#functools.lru_cache
def numSquares(self, n):
dp = self.dp
while len(dp) <= n:
dp += min(dp[-i * i] for i in range(1, int(len(dp) ** 0.5 + 1))) + 1,
return dp[n]
Here are LeetCode's official solutions with comments:
Java: DP
class Solution {
public int numSquares(int n) {
int dp[] = new int[n + 1];
Arrays.fill(dp, Integer.MAX_VALUE);
// bottom case
dp[0] = 0;
// pre-calculate the square numbers.
int max_square_index = (int) Math.sqrt(n) + 1;
int square_nums[] = new int[max_square_index];
for (int i = 1; i < max_square_index; ++i) {
square_nums[i] = i * i;
}
for (int i = 1; i <= n; ++i) {
for (int s = 1; s < max_square_index; ++s) {
if (i < square_nums[s])
break;
dp[i] = Math.min(dp[i], dp[i - square_nums[s]] + 1);
}
}
return dp[n];
}
}
Java: Greedy
class Solution {
Set<Integer> square_nums = new HashSet<Integer>();
protected boolean is_divided_by(int n, int count) {
if (count == 1) {
return square_nums.contains(n);
}
for (Integer square : square_nums) {
if (is_divided_by(n - square, count - 1)) {
return true;
}
}
return false;
}
public int numSquares(int n) {
this.square_nums.clear();
for (int i = 1; i * i <= n; ++i) {
this.square_nums.add(i * i);
}
int count = 1;
for (; count <= n; ++count) {
if (is_divided_by(n, count))
return count;
}
return count;
}
}
Java: Breadth First Search
class Solution {
public int numSquares(int n) {
ArrayList<Integer> square_nums = new ArrayList<Integer>();
for (int i = 1; i * i <= n; ++i) {
square_nums.add(i * i);
}
Set<Integer> queue = new HashSet<Integer>();
queue.add(n);
int level = 0;
while (queue.size() > 0) {
level += 1;
Set<Integer> next_queue = new HashSet<Integer>();
for (Integer remainder : queue) {
for (Integer square : square_nums) {
if (remainder.equals(square)) {
return level;
} else if (remainder < square) {
break;
} else {
next_queue.add(remainder - square);
}
}
}
queue = next_queue;
}
return level;
}
}
Java: Most efficient solution using math
Runtime: O(N ^ 0.5)
Memory: O(1)
class Solution {
protected boolean isSquare(int n) {
int sq = (int) Math.sqrt(n);
return n == sq * sq;
}
public int numSquares(int n) {
// four-square and three-square theorems.
while (n % 4 == 0)
n /= 4;
if (n % 8 == 7)
return 4;
if (this.isSquare(n))
return 1;
// enumeration to check if the number can be decomposed into sum of two squares.
for (int i = 1; i * i <= n; ++i) {
if (this.isSquare(n - i * i))
return 2;
}
// bottom case of three-square theorem.
return 3;
}
}
I'm trying to track the number of key comparisons in this binary search code. The comparison result should be around 17 for an array size of 2^16. Can you explain why I'm getting 33?
public class AdvancedBinarySearch {
int count = 0;
// Returns location of key, or -1 if not found
int AdvancedBinarySearch(int arr[], int l, int r, int x) {
int m;
while (r - l > 1) {
count++;
m = l + (r - l) / 2;
if (arr[m] <= x) {
l = m;
count++;
} else {
r = m;
count++;
}
}
if (arr[l] == x) {
count++;
return l;
}
if (arr[r] == x) {
count++;
return r;
} else {
count++;
return -1;
}
}
public void setCount(int i) {
this.count = i;
}
}
Your code is counting the if (arr[m] <= x) comparison twice. if you remove the count++ from below l = m and also from below r = m, this will no longer happen.
I have tested this before and after this change with a search for 189 in an array of the integers from 0 to 65535. This array had a size of 2^16 and before the change, 33 comparisons were counted and after the change, 17 comparisons were counted, so I think this change does what you want it to do.
I am trying to implement Binary Search Java version. From wikipedia http://en.wikipedia.org/wiki/Binary_search_algorithm#Deferred_detection_of_equality I noticed that deferred detection of equality version. It's working using that algorithm. However, when I was trying to change the if condition expression like this:
public int bsearch1(int[] numbers, int key, int start){
int L = start, R = numbers.length - 1;
while(L < R){
//find the mid point value
int mid = (L + R) / 2;
if (numbers[mid] >key){
//move to left
R = mid - 1;
} else{
// move to right, here numbers[mid] <= key
L = mid;
}
}
return (L == R && numbers[L] == key) ? L : -1;
}
It's not working properly, which goes into an infinity loop. Do you guys have any ideas about it? Thank you so much.
You've missed the effect of the assert in the Wiki you link to.
It states:
code must guarantee the interval is reduced at each iteration
You must exit if your mid >= R.
Added
The Wiki is actually a little misleading as it suggests that merely ensuring mid < r is sufficient - it is not. You must also guard against mid == min (say you have a 4 entry array and l = 2 and r = 3, mid would become 2 and stick there because 2 + 3 = 5 and 5 / 2 = 2 in integer maths).
The solution is to round up after the / 2 which can be easily achieved by:
int mid = (l + r + 1) / 2;
The final corrected and tidied code goes a little like this:
public int binarySearch(int[] numbers, int key, int start) {
int l = start, r = numbers.length - 1;
while (l < r) {
//find the mid point value
int mid = (l + r + 1) / 2;
if (numbers[mid] > key) {
//move to left
r = mid - 1;
} else {
// move to right, here numbers[mid] <= key
l = mid;
}
}
return (l == r && numbers[l] == key) ? l : -1;
}
public void test() {
int[] numbers = new int[]{1, 2, 5, 6};
for (int i = 0; i < 9; i++) {
System.out.println("Searching for " + i);
System.out.println("Found at " + binarySearch(numbers, i, 0));
}
}
There is a trivially similar algorithm here that suggests the correct approach looks more like:
public int binarySearch(int[] numbers, int key) {
int low = 0, high = numbers.length;
while (low < high) {
int mid = (low + high) / 2;
if (numbers[mid] < key) {
low = mid + 1;
} else {
high = mid;
}
}
return low < numbers.length && numbers[low] == key ? low : -1;
}
This takes a slightly different approach to the boundary conditions where high = max + 1 and also works perfectly.
You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations:
modify the i-th element in the sequence or for given x y print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.
Problem Link
I am using Segment Tree for this but i am not getting the correct output , please Help me where i have committed the mistake
CODE:
Making a Tree:
public static void maketree(int current , int a , int b ,int[] arr){
if(b<a) return;
if(b==a) {dp[current] = arr[a]; return ;}
maketree(2*current, a, (a+b)/2, arr);
maketree(2*current+1,1+ (a+b)/2, b, arr);
if(dp[2*current]>0 && dp[2*current+1]>0) dp[current] = dp[2*current] + dp[2*current+1];
else if(dp[2*current]>dp[2*current+1]) dp[current] = dp[2*current];
else dp[current] = dp[2*current+1];
}
Updating Function
public static void update(int current , int a , int b , int c , int value){
if(a>b || c<a || c>b) return ;
if(a==b){ dp[current] = value; return ; }
update(2*current, a, (a+b)/2, c, value);
update(2*current+1, (b+a)/2 +1, b, c, value);
if(dp[2*current]>0 && dp[2*current+1]>0) dp[current] = dp[2*current] + dp[2*current+1];
else if(dp[2*current]>dp[2*current+1]) dp[current] = dp[2*current];
else dp[current] = dp[2*current+1];
}
Query Function:
public static int query(int current , int a , int b , int i , int j){
int ans =0;
if(a>j || b<i || a>b) return Integer.MIN_VALUE;
if(a>=i && b<=j) return dp[current];
int x = query(2*current, a, (a+b)/2, i, j);
int y = query(2*current+1, (a+b)/2 +1, b, i, j);
if(x>0 && y>0) ans= x+y;
else if(x>y) ans = x;
else ans =y;
return ans;
}
I don;t know where i have made mistake please help , What will storage capacity required for dp array i.e. size of dp
when you are merging two nodes,then it may be like given below.execute any simple example so that you can feel it :)
void merge(node a , node b)
{
sum = a.sum + b.sum;
pre = max(a.pre , (a.sum + b.pre));
suf = max(b.suf , (b.sum + a.suf));
result = max(a.suf + b.pre,max(a.result , b.result));
}
it is quite overcomplicated imo...
int tree[1 << 17]; // 2 ^ 17 >= N * 2
int M = 1; //base of tree or sth i dont remember english name
int query(int L, int R){
int res = -10000; //minimum possible value in array
L += M - 1;
R += M - 1;
while(L <= R){
if(L % 2 == 1) res = max(res, tree[L++];
if(R % 2 == 0) res = max(res, tree[R++];
L /= 2;
R /= 2;
}
return res;
}
void update(int v, int value){
v += M - 1;
tree[v] = value;
while(v > 0){
v /= 2;
tree[v] = max(tree[v * 2], tree[v * 2 + 1]);
}
}
void make_tree(){
int n;
cin >> n;
while(M < n) M *= 2; // M is half of the size of tree
for(int i = 0;i < n;i++)
cin >> tree[i + M]; // just reading input to tree;
for(int i = M - 1;i > 0;i--) // first update for all nodes other than leafs
tree[i] = max(tree[i * 2], tree[i * 2 + 1]);
}