Create Find minimum time to finish all jobs with given constraints

Find minimum time to finish all jobs with given constraints

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+
+class solution
+{
+
+	static int getMax(int arr[], int n)
+	{
+	int result = arr[0];
+	for (int i=1; i<n; i++)
+		if (arr[i] > result)
+			result = arr[i];
+		return result;
+	}
+
+
+	static boolean isPossible(int time, int K, 
+									int job[], int n)
+	{
+
+		int cnt = 1;
+
+		int curr_time = 0; 
+
+		for (int i = 0; i < n;)
+		{
+
+			if (curr_time + job[i] > time) {
+				curr_time = 0;
+				cnt++;
+			}
+
+			// Else add time of job to current 
+			// time and move to next job.
+			else
+			{
+				curr_time += job[i];
+				i++;
+			}
+		}
+
+		// Returns true if count
+		// is smaller than k
+		return (cnt <= K);
+	}
+
+	// Returns minimum time required to 
+	// finish given array of jobs
+	// k - number of assignees
+	// T - Time required by every assignee to finish 1 unit
+	// m - Number of jobs
+	static int findMinTime(int K, int T, int job[], int n)
+	{
+		// Set start and end for binary search
+		// end provides an upper limit on time
+		int end = 0, start = 0;
+		for (int i = 0; i < n; ++i)
+			end += job[i];
+
+		// Initialize answer
+		int ans = end; 
+
+		// Find the job that takes maximum time
+		int job_max = getMax(job, n);
+
+		// Do binary search for 
+		// minimum feasible time
+		while (start <= end)
+		{
+			int mid = (start + end) / 2;
+
+			// If it is possible to finish jobs in mid time
+			if (mid >= job_max && isPossible(mid, K, job, n))
+			{
+				ans = Math.min(ans, mid); 
+
+				end = mid - 1;
+			}
+
+			else
+				start = mid + 1;
+		}
+
+		return (ans * T);
+	}
+
+	public static void main(String arg[])
+	{
+		int job[] = {10, 7, 8, 12, 6, 8};
+		int n = job.length;
+		int k = 4, T = 5;
+		System.out.println(findMinTime(k, T, job, n));
+	}
+}
+