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13 paź 2024 · Given an array arr [], the task is to find all possible indices (i, j) of pairs (arr [i], arr [j]) whose sum is equal to 0 and i != j. We can return pairs in any order, but all the returned pairs should be internally sorted, that is for any pair(i, j), i should be less than j. Examples: Explanation: There are no pairs with zero sum.
31 sie 2024 · void twoSumTwoPointer (int arr [], int n, int target) {int left = 0, right = n-1; // Iterate while left pointer is less than right while (left < right) {int sum = arr [left] + arr [right]; // Check if the sum matches the target if (sum == target) {printf ("Pair found: %d and %d \n ", arr [left], arr [right]); return;} else if (sum < target ...
24 lis 2014 · public static int sumOfDigits(int n) { String digits = new Integer(n).toString(); int sum = 0; for (char c: digits.toCharArray()) sum += c - '0'; return sum; } You can use it like this: System.out.printf("Sum of digits = %d%n", sumOfDigits(321));
Two Number Sum Problem solution in Java METHOD 1. Naive approach: Use two for loops. The naive approach is to just use two nested for loops and check if the sum of any two elements in the array is equal to the given target. Time complexity: O(n^2)
22 paź 2024 · Given an array arr [] of n integers and a target value, the task is to find whether there is a pair of elements in the array whose sum is equal to target. This problem is a variation of 2Sum problem. Examples: There is no pair with sum equals to given target.
10 mar 2018 · public int[] twoSum(int[] nums, int target) { int complement; . //loop to check every element in the array. for (int x = 0; x<nums.length; x++) { . complement = target - nums[x]; //loop to...
This lesson introduces the two-pointer technique in Java for efficiently finding pairs in an array that sum up to a given target value. It covers the naive approach, explains the improved two-pointer method, and walks through step-by-step solution building with complexity analysis.
