Implement Solution for Painter's Partition Problem

C++/paintersPartition.cpp

@@ -0,0 +1,62 @@
+/*
+Given n boards of different lengths represented by an array arr of size n, and m painters,
+ you need to determine the minimum amount of time required to paint all the boards. Each painter paints a contiguous segment of boards.
+ Constraints:
+
+A painter can only paint continuous segments of boards.
+Each painter takes the same amount of time to paint a unit length of the board.
+You have to minimize the maximum time taken by a painter to paint their assigned boards.
+Input:
+An integer n representing the number of boards.
+An integer m representing the number of painters.
+An array arr[] of size n, where each element arr[i] represents the length of the i-th board.
+Output:
+Return the minimum time required to paint all the boards such that no painter is overloaded with work.
+*/
+
+#include<iostream>
+#include<vector>
+#include <climits>
+using namespace std;
+
+bool isPossible(vector<int>&arr, int n, int m, int maxAllowedTime){
+  int painter = 1, time = 0;
+  for(int i=0; i<n; i++){
+    if(time+arr[i]<= maxAllowedTime){
+        time += arr[i];
+    } else{
+        painter++;
+        time = arr[i];
+
+    }
+  }
+  return painter<=m;
+}
+int minTimeToPaint(vector<int> &arr, int n, int m){
+    int sum =0, maxVal = INT_MIN;
+    for(int i=0; i<arr.size(); i++){
+        sum += arr[i];
+        maxVal = max(maxVal, arr[i]);
+    }
+
+    int st = maxVal, end = sum, ans = -1;
+    while(st<=end){
+        int mid = st + (end-st)/2;
+        if(isPossible(arr, n, m, mid)){
+            ans = mid;
+            end = mid-1;
+        }
+        else{
+            st = mid+1;
+        }
+
+    }
+    return ans;
+}
+
+int main(){
+    vector<int>arr = {40,30,10,20};
+    int n = 4, m =2;
+    cout<<minTimeToPaint(arr, n, m)<<endl;
+    return 0;
+}