JAVA SOLUTION EASY and EXPLAINED
JAVA SOLUTION @DeepakKumar\n# In Case of Any Doubt Feel Free to ASK .....\n\n\nclass Solution {\n    public int maximumPopulation(int[][] logs) {\n        // De
# JAVA SOLUTION @DeepakKumar\n# In Case of Any Doubt Feel Free to ASK .....\n\n```\nclass Solution {\n    public int maximumPopulation(int[][] logs) {\n        // Declaring Array of Size 101\n        int [] arr = new int[101];  // Initializing of 101 BECAUSE OF Constraint 1950 <= birthi < deathi <= 2050\n        for (int[] log : logs) {    // For Every row in logs\n            for (int j = log[0]; j < log[1]; j++) {   // Range is from [birthi, deathi - 1] Both Inclusive\n                arr[j - 1950]++;        \n                /*\n                During j = 1950 \n                arr[1950-1950] = arr[0] will be incremented\n                Similarly During j = 1951\n                arr[1] will be incremented ..... and so on..\n                */\n            }\n        }\n       // System.out.println(Arrays.toString(arr));\n        int maxValue = 0;     // Initialize maxValue of Population as 0\n        int maxYear = 1950;   // Let Initially YEAR with Maximum Population is 1950\n        for(int i=0;i<arr.length;i++){   // Now compare maxValue with all the Elements of arr\n            if(arr[i]> maxValue){        // If any Element is Greater than it\n                maxValue = arr[i];       // update maxValue\n                maxYear = i+1950;       // Once maxValue is Incremented then Also Increment the year \n                /*\n                 EXAMPLE \n                    Suppose i = 2 has MAX VALUE\n                    so YEAR is 2+1950 = 1952\n                */\n            }\n        }\n        return maxYear;\n    }\n}\n```
['Java']