Algorithm
Problem Name: 747. Largest Number At Least Twice of Others
You are given an integer array nums where the largest integer is unique.
Determine whether the largest element in the array is at least twice as much as every other number in the array. If it is, return the index of the largest element, or return -1 otherwise.
Example 1:
Input: nums = [3,6,1,0] Output: 1 Explanation: 6 is the largest integer. For every other number in the array x, 6 is at least twice as big as x. The index of value 6 is 1, so we return 1.
Example 2:
Input: nums = [1,2,3,4] Output: -1 Explanation: 4 is less than twice the value of 3, so we return -1.
Constraints:
2 <= nums.length <= 500 <= nums[i] <= 100- The largest element in
numsis unique.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
int dominantIndex(vector<int>& nums) {
int a = 0, b = 0, idx = -1;
for(int i = 0; i < nums.size(); i++){
if(nums[i] > a){
b = a;
a = nums[i];
idx = i;
}
else if(nums[i] > b) b = nums[i];
}
return a >= 2 * b ? idx : -1;
}
};
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int dominantIndex(int[] nums) {
int maxNum = Integer.MIN_VALUE;
int secMaxNum = Integer.MIN_VALUE;
int maxIdx = -1;
for (int i = 0; i < nums.length; i++) {
if (nums[i] >= maxNum) {
secMaxNum = maxNum;
maxNum = nums[i];
maxIdx = i;
}
else if (nums[i] > secMaxNum) {
secMaxNum = nums[i];
}
}
return maxNum >= 2 * secMaxNum ? maxIdx : -1;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const dominantIndex = function(nums) {
let max = -Infinity
let idx = -1
for(let i = 0, len = nums.length; i < len; i++) {
if(nums[i] > max) {
max = nums[i]
idx = i
}
}
for(let i = 0, len = nums.length; i < len; i++) {
if(nums[i] !== max && max < nums[i] * 2> return -1
}
return idx
};
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def countSubstrings(self, s):
res = 0
for k in range(len(s)):
i = j = k
while 0 <= i and j < len(s):
if s[i] == s[j]: res += 1
else: break
i , j = i - 1, j + 1
i , j =k , k + 1
while 0 <= i and j < len(s):
if s[i] == s[j]: res += 1
else: break
i , j = i - 1, j + 1
return res
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Input
Output
#5 Code Example with C# Programming
Code -
C# Programming
namespace LeetCode
{
public class _0747_LargestNumberAtLeastTwiceOfOthers
{
public int DominantIndex(int[] nums)
{
int max1 = int.MinValue, max2 = int.MinValue;
int index = -1;
for (int i = 0; i < nums.Length; i++)
{
var num = nums[i];
if (num > max2)
if (num >= max1)
{
max2 = max1;
max1 = num;
index = i;
}
else
{
max2 = num;
}
}
return max1 >= 2 * max2 ? index : -1;
}
}
}
Input
Output