Day - 79 work

Author: mandliya

README.md

@@ -6,9 +6,9 @@
 
 | Current Status|     Stats     |
 | :------------: | :----------: |
-| Total Problems | 114 |
-| Current Streak | 78 days |
-| Longest Streak | 78 ( August 17, 2015 - November 02, 2015 ) |
+| Total Problems | 116 |
+| Current Streak | 79 days |
+| Longest Streak | 79 ( August 17, 2015 - November 03, 2015 ) |
 
 </center>
 
@@ -157,6 +157,7 @@ Include contains single header implementation of data structures and some algori
 | Given a list of unsorted integers, A={a<sub>1</sub>,a<sub>2</sub>,…,a<sub>N</sub>}, Find the pair of elements that have the smallest absolute difference between them? If there are multiple pairs, find them all.| [closest_numbers.cpp](sort_search_problems/closest_numbers.cpp)|
 | Given a sorted array, determine index of fixed point in this array. If array does not have a fixed point return -1. An array has a fixed point when index of the element is same as index i.e. i == arr[i], Expected time complexity O(logn)| [fixedPoint.cpp](sort_search_problems/fixedPoint.cpp)|
 | Find the maximum element in an array which is first increasing and then decreasing. Input: arr[] = {8, 10, 20, 80, 100, 200, 400, 500, 3, 2, 1}, output : 500. Array may be strictly increasing or decreasing as well. ExpectedTime complexity is O(logn).| [findMaximum.cpp](sort_search_problems/findMaximum.cpp)|
+| Given an array of positive and/or negative integers, find a pair in the array whose sum is closest to 0.| [findClosestPairToZero.cpp](sort_search_problems/findClosestPairToZero.cpp)|
 
 ### Graph Problems
 | Problem | Solution |
@@ -191,3 +192,4 @@ Include contains single header implementation of data structures and some algori
 |Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is *"((()))", "(()())", "(())()", "()(())", "()()()"*| [generate_parenthesis.cpp](leet_code_problems/generate_parenthesis.cpp)|
 |Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.For example, Given nums = [0, 1, 3] return 2.| [missing_number.cpp](leet_code_problems/missing_number.cpp)|
 |Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array.| [find_min_rotated.cpp](leet_code_problems/find_min_rotated.cpp)|
+|Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.| [threeSumClosest.cpp](leet_code_problems/threeSumClosest.cpp)|

leet_code_problems/run

@@ -0,0 +1,47 @@
+/**
+ * Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target.
+ * Return the sum of the three integers. You may assume that each input would have exactly one solution.
+ * For example, given array S = {-1 2 1 -4}, and target = 1.
+ * The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
+ */
+
+
+#include <iostream>
+#include <vector>
+#include <algorithm>
+
+
+int threeSumClosest(std::vector<int>& nums, int target) {
+	int closestSum = (nums[0] + nums[1] + nums[2]);
+	std::sort(nums.begin(), nums.end());
+	for ( size_t i = 0 ;i < nums.size() - 2; ++i ) {
+		size_t k = nums.size() - 1;
+		size_t j = i + 1;
+		while( j < k )  {
+			int currSum = (nums[i] + nums[j] + nums[k]);
+			if ( currSum == target ) {
+				return currSum;
+			}
+			if ( abs(target - closestSum) > abs(target - currSum)) {
+				closestSum = currSum;
+			}
+			if ( currSum > target ) {
+				--k;
+			} else {
+				++j;
+			}
+		}
+	}
+	return closestSum;
+}
+
+int main() {
+	std::vector<int> vec{ -1,2,1,-4 };
+	std::cout << "Vec:";
+	for ( auto c : vec ) {
+		std::cout << c << " ";
+	}
+	std::cout << std::endl;
+	std::cout << "Three sum closest to 1 in above vector is :" << threeSumClosest(vec, 1) << std::endl;
+	return 0;
+}

leet_code_problems/threeSumClosest.cpp

@@ -0,0 +1,59 @@
+/**
+ * An Array of integers is given, both +ve and -ve. You need to find the two elements such that their sum is closest to zero.
+ * Example:
+ * vec : { 1, 60, -10, 70, -80, 85 }
+ *
+ * output: -80, 85
+ *
+ * If no such pair exists return a { 0, 0 } pair
+ *
+ */
+
+#include <iostream>
+#include <vector>
+#include <algorithm>
+#include <utility>
+#include <limits>
+
+std::pair<int, int> pairSumClosestToZero( std::vector<int> & vec ) {
+	std::pair<int, int> minPair( 0, 0 );
+	if ( vec.size() < 2 ) {
+		return minPair;
+	}
+
+	size_t minLeft = 0, minRight = vec.size()-1;
+	size_t left = 0, right = vec.size() -1;
+	int minSum = std::numeric_limits<int>::max();
+
+	std::sort( vec.begin(), vec.end() );
+
+	while( left < right ) {
+		int sum = vec[left] + vec[right];
+		if ( std::abs(minSum) > std::abs(sum) ) {
+			minSum = sum;
+			minLeft = left;
+			minRight = right;
+		}
+		if ( sum > 0 ) {
+			--right;
+		} else {
+			++left;
+		}
+	}
+	minPair.first = vec[minLeft];
+	minPair.second = vec[minRight];
+	return minPair;
+}
+
+
+int main() {
+	std::vector<int> vec{  1, 60, -10, 70, -80, 85 };
+	std::cout << "Vec: ";
+	for ( auto v : vec ) {
+		std::cout << v << " ";
+	}
+	std::pair<int, int> minPair = pairSumClosestToZero(vec);
+	std::cout << "Pair whose sum is closes to 0 in above vector is: ("
+			  << minPair.first << ", " << minPair.second << ")" << std::endl;
+	return 0;
+}