Given an array stalls[] which denotes the position of a stall and an integer k which denotes the number of aggressive cows. The task is to assign stalls to k cows such that the minimum distance between any two of them is the maximum possible.
Examples:
Input: stalls[] = [1, 2, 4, 8, 9], k = 3
Output: 3
Explanation: We can place cow 1 at position 1, cow 2 at position 4 and cow 3 at position 9. So, the maximum possible minimum distance between two cows is 3.
Input: stalls[] = [6, 7, 9, 11, 13, 15], k = 4
Output: 2
Explanation: We can place cow 1 at position 6, cow 2 at position 9, cow 3 at position 11 and cow 4 at position 15. So, the maximum possible minimum distance between two cows is 2.
[Naive Approach] By iterating over all possible distances
The idea is to iterate over all possible distances starting from 1 up to the difference between the farthest stalls. First, we sort the array to ensure the stalls are in the correct sequence. Then, for each distance, we try to place the cows greedily - placing the first cow in the first stall and the next ones only if the gap from the last placed cow is at least the current distance. If all cows can be placed for a certain distance, we update our result. The process continues until all distances are checked.
C++
// C++ program to find maximum possible minimum distance
// between any two cows by iterating over all distances
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// function to check if we can place k cows
// with at least dist distance apart
bool check(vector<int> &stalls, int k, int dist) {
// Place first cow at 0th index
int cnt = 1;
int prev = stalls[0];
for (int i = 1; i < stalls.size(); i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
int aggressiveCows(vector<int> &stalls, int k) {
// sorting the array to ensure stalls in sequence
sort(stalls.begin(), stalls.end());
int res = 0;
// Minimum and maximum possible minimum distance
// between two stalls
int minDist = 1;
int maxDist = stalls.back() - stalls[0];
// Iterating through all possible distances
for (int i = minDist; i <= maxDist; i++) {
// If we can place k cows with the
// current distance i, update the res
if (check(stalls, k, i))
res = i;
}
return res;
}
int main() {
vector<int> stalls = {1, 2, 4, 8, 9};
int k = 3;
int ans = aggressiveCows(stalls, k);
cout << ans;
return 0;
}
// C program to find maximum possible minimum distance
// between any two cows by iterating over all distances
#include <stdio.h>
#include <stdlib.h>
int compare(const void *a, const void *b) {
return (*(int*)a - *(int*)b);
}
// function to check if we can place k cows
// with at least dist distance apart
int check(int stalls[], int size, int k, int dist) {
// Place first cow at 0th index
int cnt = 1;
int prev = stalls[0];
for (int i = 1; i < size; i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
int aggressiveCows(int stalls[], int size, int k) {
// sorting the array to ensure stalls in sequence
qsort(stalls, size, sizeof(int), (int (*)(const void *, const void *))compare);
int res = 0;
// Minimum and maximum possible minimum distance
// between two stalls
int minDist = 1;
int maxDist = stalls[size - 1] - stalls[0];
// Iterating through all possible distances
for (int i = minDist; i <= maxDist; i++) {
// If we can place k cows with the
// current distance i, update the res
if (check(stalls, size, k, i))
res = i;
}
return res;
}
int main() {
int stalls[] = {1, 2, 4, 8, 9};
int k = 3;
int size = sizeof(stalls) / sizeof(stalls[0]);
int ans = aggressiveCows(stalls, size, k);
printf("%d\n", ans);
return 0;
}
// Java program to find maximum possible minimum distance
// between any two cows by iterating over all distances
import java.util.Arrays;
class GfG {
// function to check if we can place k cows
// with at least dist distance apart
static boolean check(int[] stalls, int k, int dist) {
// Place first cow at 0th index
int cnt = 1;
int prev = stalls[0];
for (int i = 1; i < stalls.length; i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
static int aggressiveCows(int[] stalls, int k) {
// sorting the array to ensure stalls in sequence
Arrays.sort(stalls);
int res = 0;
// Minimum and maximum possible minimum distance
// between two stalls
int minDist = 1;
int maxDist = stalls[stalls.length - 1] - stalls[0];
// Iterating through all possible distances
for (int i = minDist; i <= maxDist; i++) {
// If we can place k cows with the
// current distance i, update the res
if (check(stalls, k, i))
res = i;
}
return res;
}
public static void main(String[] args) {
int[] stalls = {1, 2, 4, 8, 9};
int k = 3;
int ans = aggressiveCows(stalls, k);
System.out.println(ans);
}
}
# Python program to find maximum possible minimum distance
# between any two cows by iterating over all distances
# function to check if we can place k cows
# with at least dist distance apart
def check(stalls, k, dist):
# Place first cow at 0th index
cnt = 1
prev = stalls[0]
for i in range(1, len(stalls)):
# If the current stall is at least dist away
# from the previous one place the cow here
if stalls[i] - prev >= dist:
prev = stalls[i]
cnt += 1
# Return true if we are able to place all 'k' cows
return cnt >= k
def aggressiveCows(stalls, k):
# sorting the array to ensure stalls in sequence
stalls.sort()
res = 0
# Minimum and maximum possible minimum distance
# between two stalls
minDist = 1
maxDist = stalls[-1] - stalls[0]
# Iterating through all possible distances
for i in range(minDist, maxDist + 1):
# If we can place k cows with the
# current distance i, update the res
if check(stalls, k, i):
res = i
return res
if __name__ == "__main__":
stalls = [1, 2, 4, 8, 9]
k = 3
ans = aggressiveCows(stalls, k)
print(ans)
// C# program to find maximum possible minimum distance
// between any two cows by iterating over all distances
using System;
class GfG {
// function to check if we can place k cows
// with at least dist distance apart
static bool check(int[] stalls, int k, int dist) {
// Place first cow at 0th index
int cnt = 1;
int prev = stalls[0];
for (int i = 1; i < stalls.Length; i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
static int aggressiveCows(int[] stalls, int k) {
// sorting the array to ensure stalls in sequence
Array.Sort(stalls);
int res = 0;
// Minimum and maximum possible minimum distance
// between two stalls
int minDist = 1;
int maxDist = stalls[stalls.Length - 1] - stalls[0];
// Iterating through all possible distances
for (int i = minDist; i <= maxDist; i++) {
// If we can place k cows with the
// current distance i, update the res
if (check(stalls, k, i))
res = i;
}
return res;
}
static void Main() {
int[] stalls = {1, 2, 4, 8, 9};
int k = 3;
int ans = aggressiveCows(stalls, k);
Console.WriteLine(ans);
}
}
// JavaScript program to find maximum possible minimum distance
// between any two cows by iterating over all distances
// function to check if we can place k cows
// with at least dist distance apart
function check(stalls, k, dist) {
// Place first cow at 0th index
let cnt = 1;
let prev = stalls[0];
for (let i = 1; i < stalls.length; i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
function aggressiveCows(stalls, k) {
// sorting the array to ensure stalls in sequence
stalls.sort((a, b) => a - b);
let res = 0;
// Minimum and maximum possible minimum distance
// between two stalls
let minDist = 1;
let maxDist = stalls[stalls.length - 1] - stalls[0];
// Iterating through all possible distances
for (let i = minDist; i <= maxDist; i++) {
// If we can place k cows with the
// current distance i, update the res
if (check(stalls, k, i))
res = i;
}
return res;
}
let stalls = [1, 2, 4, 8, 9];
let k = 3;
let ans = aggressiveCows(stalls, k);
console.log(ans);
Time Complexity: O(n*(MAX - MIN)), where n is the size of the array, MAX is the maximum element in the array and MIN is minimum element in the array.
Auxiliary Space: O(1).
[Expected Approach] Using Binary Search
The minimum distance between the cows has a monotonic property.
- If we can place all the cows with a minimum distance d, then we can also place them with any smaller distance than d. The reason is by reducing the minimum gap, we can place more number of cows as lower gaps allow us to place cows even more closer.
- If we can't place all the cows with a minimum distance d, then we can't place them with any larger distance than d. The reason is if the gap is already too large to place the cows, then larger gap will also not work.
Therefore, we can use binary search to find the maximum possible minimum distance, and to check the for any distance, we place the first cow in the first stall and the next ones only if the gap from the last placed cow is at least that distance. If all cows can be placed for a certain distance, then it is a feasible distance.
C++
// C++ program to find maximum possible minimum distance
// between any two cows using binary search
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// function to check if we can place k cows
// with at least dist distance apart
bool check(vector<int> &stalls, int k, int dist) {
// Place first cow at 0th index
int cnt = 1;
int prev = stalls[0];
for (int i = 1; i < stalls.size(); i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
int aggressiveCows(vector<int> &stalls, int k) {
// sorting the array to ensure stalls in sequence
sort(stalls.begin(), stalls.end());
int res = 0;
// Search Space for Binary Search
int lo = 1;
int hi = stalls.back() - stalls[0];
while(lo <= hi) {
int mid = lo + (hi - lo)/2;
// If the mid ditance is possible, update
// the result and search for larger ditance
if(check(stalls, k, mid)) {
res = mid;
lo = mid + 1;
}
else {
hi = mid - 1;
}
}
return res;
}
int main() {
vector<int> stalls = {1, 2, 4, 8, 9};
int k = 3;
int ans = aggressiveCows(stalls, k);
cout << ans;
return 0;
}
// C program to find maximum possible minimum distance
// between any two cows using binary search
#include <stdio.h>
int compare(const void *a, const void *b) {
return (*(int*)a - *(int*)b);
}
// function to check if we can place k cows
// with at least dist distance apart
int check(int stalls[], int n, int k, int dist) {
// Place first cow at 0th index
int cnt = 1;
int prev = stalls[0];
for (int i = 1; i < n; i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
int aggressiveCows(int stalls[], int n, int k) {
// sorting the array to ensure stalls in sequence
qsort(stalls, n, sizeof(int), compare);
int res = 0;
// Search Space for Binary Search
int lo = 1;
int hi = stalls[n - 1] - stalls[0];
while(lo <= hi) {
int mid = lo + (hi - lo) / 2;
// If the mid distance is possible, update
// the result and search for larger distance
if(check(stalls, n, k, mid)) {
res = mid;
lo = mid + 1;
}
else {
hi = mid - 1;
}
}
return res;
}
int main() {
int stalls[] = {1, 2, 4, 8, 9};
int k = 3;
int n = sizeof(stalls) / sizeof(stalls[0]);
int ans = aggressiveCows(stalls, n, k);
printf("%d\n", ans);
return 0;
}
// Java program to find maximum possible minimum distance
// between any two cows using binary search
import java.util.Arrays;
class GfG {
// function to check if we can place k cows
// with at least dist distance apart
static boolean check(int[] stalls, int k, int dist) {
// Place first cow at 0th index
int cnt = 1;
int prev = stalls[0];
for (int i = 1; i < stalls.length; i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
static int aggressiveCows(int[] stalls, int k) {
// sorting the array to ensure stalls in sequence
Arrays.sort(stalls);
int res = 0;
// Search Space for Binary Search
int lo = 1;
int hi = stalls[stalls.length - 1] - stalls[0];
while(lo <= hi) {
int mid = lo + (hi - lo) / 2;
// If the mid distance is possible, update
// the result and search for larger distance
if(check(stalls, k, mid)) {
res = mid;
lo = mid + 1;
}
else {
hi = mid - 1;
}
}
return res;
}
public static void main(String[] args) {
int[] stalls = {1, 2, 4, 8, 9};
int k = 3;
int ans = aggressiveCows(stalls, k);
System.out.println(ans);
}
}
# Python program to find maximum possible minimum distance
# between any two cows using binary search
def check(stalls, k, dist):
# Place first cow at 0th index
cnt = 1
prev = stalls[0]
for i in range(1, len(stalls)):
# If the current stall is at least dist away
# from the previous one place the cow here
if stalls[i] - prev >= dist:
prev = stalls[i]
cnt += 1
# Return true if we are able to place all 'k' cows
return cnt >= k
def aggressiveCows(stalls, k):
# sorting the array to ensure stalls in sequence
stalls.sort()
res = 0
# Search Space for Binary Search
lo = 1
hi = stalls[-1] - stalls[0]
while lo <= hi:
mid = lo + (hi - lo) // 2
# If the mid distance is possible, update
# the result and search for larger distance
if check(stalls, k, mid):
res = mid
lo = mid + 1
else:
hi = mid - 1
return res
if __name__ == "__main__":
stalls = [1, 2, 4, 8, 9]
k = 3
ans = aggressiveCows(stalls, k)
print(ans)
// C# program to find maximum possible minimum distance
// between any two cows using binary search
using System;
class GfG {
// function to check if we can place k cows
// with at least dist distance apart
static bool check(int[] stalls, int k, int dist) {
// Place first cow at 0th index
int cnt = 1;
int prev = stalls[0];
for (int i = 1; i < stalls.Length; i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
static int aggressiveCows(int[] stalls, int k) {
// sorting the array to ensure stalls in sequence
Array.Sort(stalls);
int res = 0;
// Search Space for Binary Search
int lo = 1;
int hi = stalls[stalls.Length - 1] - stalls[0];
while(lo <= hi) {
int mid = lo + (hi - lo) / 2;
// If the mid distance is possible, update
// the result and search for larger distance
if(check(stalls, k, mid)) {
res = mid;
lo = mid + 1;
}
else {
hi = mid - 1;
}
}
return res;
}
static void Main() {
int[] stalls = {1, 2, 4, 8, 9};
int k = 3;
int ans = aggressiveCows(stalls, k);
Console.WriteLine(ans);
}
}
// JavaScript program to find maximum possible minimum distance
// between any two cows using binary search
function check(stalls, k, dist) {
// Place first cow at 0th index
let cnt = 1;
let prev = stalls[0];
for (let i = 1; i < stalls.length; i++) {
// If the current stall is at least dist away
// from the previous one place the cow here
if (stalls[i] - prev >= dist) {
prev = stalls[i];
cnt++;
}
}
// Return true if we are able to place all 'k' cows
return (cnt >= k);
}
function aggressiveCows(stalls, k) {
// sorting the array to ensure stalls in sequence
stalls.sort((a, b) => a - b);
let res = 0;
// Search Space for Binary Search
let lo = 1;
let hi = stalls[stalls.length - 1] - stalls[0];
while (lo <= hi) {
let mid = Math.floor(lo + (hi - lo) / 2);
// If the mid distance is possible, update
// the result and search for larger distance
if (check(stalls, k, mid)) {
res = mid;
lo = mid + 1;
} else {
hi = mid - 1;
}
}
return res;
}
const stalls = [1, 2, 4, 8, 9];
const k = 3;
const ans = aggressiveCows(stalls, k);
console.log(ans);
Time Complexity: O(n * log(MAX - MIN)) - where n is the size of the array, MAX is the maximum element in the array and MIN is minimum element in the array.
Auxiliary Space: O(1)
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