Given an array of pairs where each pair represents a range, the task is to find the sum of the minimum difference between the consecutive elements of an array where the array is filled in the below manner:
- Each element of an array lies in the range given at its corresponding index in the range array.
- Final sum of difference of consecutive elements in the array formed is minimum.
Examples:
Input: range[] = {{2, 4}, {3, 6}, {1, 5}, {1, 3}, {2, 7}}
Output: 0
The result is 0 because the array {3, 3, 3, 3, 3} is chosen
then the sum of difference of consecutive element will be
{ |3-3| + |3-3| + |3-3| + |3-3| } = 0 which the minimum.
Input: range[] = {{1, 3}, {2, 5}, {6, 8}, {1, 2}, {2, 3}}
Output: 7
The result is 7 because if the array {3, 3, 6, 2, 2} is chosen
then the sum of difference of consecutive element will be
{ |3-3| + |6-3| + |2-6| + |2-2| }= 7 which is the minimum.
Approach: A greedy approach has been followed to solve the above problem. Initially, we have to fill the array in such a way that the sum of the difference obtained is minimum. The greedy approach is as follows:
- If the range of the previous index intersects the range of current index then, in this case, the minimum difference will be 0 and store the highest value and the lowest value of intersecting range.
- If the lowest value of the range of the previous index is greater than the highest value for the current index then in this case the least possible sum is the difference in the lowest value of the previous range and the highest value stored in the current range.
- If the highest range of the previous index is lower than the lowest value of current range then the minimum sum is the difference in the lowest value stored for the current index and the highest range of the previous index.
Below is the implementation of the above approach:
C++
// C++ program for finding the minimum sum of// difference between consecutive elements#include <bits/stdc++.h>using namespace std;// function to find minimum sum of// difference of consecutive elementint solve(pair<int, int> v[], int n){ // ul to store upper limit // ll to store lower limit int ans, ul, ll; // storethe lower range in ll // and upper range in ul ll = v[0].first; ul = v[0].second; // initialize the answer with 0 ans = 0; // iterate for all ranges for (int i = 1; i < n; i++) { // case 1, in this case the difference will be 0 if ((v[i].first <= ul && v[i].first >= ll) || (v[i].second >= ll && v[i].second <= ul)) { // change upper limit and lower limit if (v[i].first > ll) { ll = v[i].first; } if (v[i].second < ul) { ul = v[i].second; } } // case 2 else if (v[i].first > ul) { // store the difference ans += abs(ul - v[i].first); ul = v[i].first; ll = v[i].first; } // case 3 else if (v[i].second < ll) { // store the difference ans += abs(ll - v[i].second); ul = v[i].second; ll = v[i].second; } } return ans;}// Driver codeint main(){ // array of range pair<int, int> v[] = { { 1, 3 }, { 2, 5 }, { 6, 8 }, { 1, 2 }, { 2, 3 } }; int n = sizeof(v) / sizeof(v[0]); cout << solve(v, n) << endl; return 0;} |
Java
// Java program for finding the // minimum sum of difference// between consecutive elementsimport java.io.*;class GFG{ // function to find minimum // sum of difference of// consecutive elementstatic int solve(int[][] v, int n){ // ul to store upper limit // ll to store lower limit int ans, ul, ll; int first = 0; int second = 1; // storethe lower range // in ll and upper range // in ul ll = v[0][first]; ul = v[0][second]; // initialize the // answer with 0 ans = 0; // iterate for all ranges for (int i = 1; i < n; i++) { // case 1, in this case // the difference will be 0 if ((v[i][first] <= ul && v[i][first] >= ll) || (v[i][second] >= ll && v[i][second] <= ul)) { // change upper limit // and lower limit if (v[i][first] > ll) { ll = v[i][first]; } if (v[i][second] < ul) { ul = v[i][second]; } } // case 2 else if (v[i][first] > ul) { // store the difference ans += Math.abs(ul - v[i][first]); ul = v[i][first]; ll = v[i][first]; } // case 3 else if (v[i][second] < ll) { // store the difference ans += Math.abs(ll - v[i][second]); ul = v[i][second]; ll = v[i][second]; } } return ans;}// Driver codepublic static void main(String []args){ // array of range int[][] v = {{ 1, 3 }, { 2, 5 }, { 6, 8 }, { 1, 2 }, { 2, 3 }}; int n = 5; System.out.println(solve(v, n));}}// This code is contributed// by chandan_jnu |
Python
# Python program for finding # the minimum sum of difference# between consecutive elementsclass pair: first = 0 second = 0 def __init__(self, a, b): self.first = a self.second = b# function to find minimum # sum of difference of # consecutive elementdef solve(v, n): # ul to store upper limit # ll to store lower limit ans = 0; ul = 0; ll = 0; # storethe lower range # in ll and upper range # in ul ll = v[0].first ul = v[0].second # initialize the # answer with 0 ans = 0 # iterate for all ranges for i in range(1, n): # case 1, in this case # the difference will be 0 if (v[i].first <= ul and v[i].first >= ll) or \ (v[i].second >= ll and v[i].second <= ul): # change upper limit # and lower limit if v[i].first > ll: ll = v[i].first if v[i].second < ul: ul = v[i].second; # case 2 elif v[i].first > ul: # store the difference ans += abs(ul - v[i].first) ul = v[i].first ll = v[i].first # case 3 elif v[i].second < ll: # store the difference ans += abs(ll - v[i].second); ul = v[i].second; ll = v[i].second; return ans# Driver code# array of rangev = [pair(1, 3), pair(2, 5), pair(6, 8), pair(1, 2), pair(2, 3) ]n = len(v)print(solve(v, n))# This code is contributed# by Harshit Saini |
C#
// C# program for finding the // minimum sum of difference// between consecutive elementsusing System;class GFG{// function to find minimum // sum of difference of// consecutive elementstatic int solve(int[,] v, int n){ // ul to store upper limit // ll to store lower limit int ans, ul, ll; int first = 0; int second = 1; // storethe lower range // in ll and upper range // in ul ll = v[0, first]; ul = v[0, second]; // initialize the // answer with 0 ans = 0; // iterate for all ranges for (int i = 1; i < n; i++) { // case 1, in this case // the difference will be 0 if ((v[i, first] <= ul && v[i, first] >= ll) || (v[i, second] >= ll && v[i, second] <= ul)) { // change upper limit // and lower limit if (v[i, first] > ll) { ll = v[i, first]; } if (v[i, second] < ul) { ul = v[i, second]; } } // case 2 else if (v[i, first] > ul) { // store the difference ans += Math.Abs(ul - v[i, first]); ul = v[i, first]; ll = v[i, first]; } // case 3 else if (v[i, second] < ll) { // store the difference ans += Math.Abs(ll - v[i, second]); ul = v[i, second]; ll = v[i, second]; } } return ans;}// Driver codestatic void Main(){ // array of range int[,] v = new int[5,2]{ { 1, 3 }, { 2, 5 }, { 6, 8 }, { 1, 2 }, { 2, 3 } }; int n = 5; Console.WriteLine(solve(v, n));}}// This code is contributed// by chandan_jnu |
PHP
<?php// PHP program for finding the // minimum sum of difference// between consecutive elements// function to find minimum // sum of difference of// consecutive elementfunction solve($v, $n){// ul to store upper limit// ll to store lower limit$ans; $ul; $ll;$first = 0;$second = 1;// storethe lower range // in ll and upper range // in ul$ll = $v[0][$first];$ul = $v[0][$second];// initialize the// answer with 0$ans = 0;// iterate for all rangesfor ($i = 1; $i <$n; $i++) { // case 1, in this case // the difference will be 0 if (($v[$i][$first] <= $ul and $v[$i][$first] >= $ll) or ($v[$i][$second] >= $ll and $v[$i][$second] <= $ul)) { // change upper limit // and lower limit if ($v[$i][$first] > $ll) { $ll = $v[$i][$first]; } if ($v[$i][$second] < $ul) { $ul = $v[$i][$second]; } } // case 2 else if ($v[$i][$first] > $ul) { // store the difference $ans += abs($ul - $v[$i][$first]); $ul = $v[$i][$first]; $ll = $v[$i][$first]; } // case 3 else if ($v[$i][$second] < $ll) { // store the difference $ans += abs($ll - $v[$i][$second]); $ul = $v[$i][$second]; $ll = $v[$i][$second]; }}return $ans;}// Driver code// array of range$v = array(array( 1, 3 ), array( 2, 5 ), array( 6, 8 ), array( 1, 2 ), array( 2, 3 ));$n = 5;echo(solve($v, $n));// This code is contributed// by chandan_jnu?> |
Javascript
<script>// JavaScript program for finding the // minimum sum of difference// between consecutive elements// function to find minimum // sum of difference of// consecutive elementfunction solve(v, n){ // ul to store upper limit // ll to store lower limit let ans, ul, ll; let first = 0; let second = 1; // storethe lower range // in ll and upper range // in ul ll = v[0][first]; ul = v[0][second]; // initialize the // answer with 0 ans = 0; // iterate for all ranges for (let i = 1; i < n; i++) { // case 1, in this case // the difference will be 0 if ((v[i][first] <= ul && v[i][first] >= ll) || (v[i][second] >= ll && v[i][second] <= ul)) { // change upper limit // and lower limit if (v[i][first] > ll) { ll = v[i][first]; } if (v[i][second] < ul) { ul = v[i][second]; } } // case 2 else if (v[i][first] > ul) { // store the difference ans += Math.abs(ul - v[i][first]); ul = v[i][first]; ll = v[i][first]; } // case 3 else if (v[i][second] < ll) { // store the difference ans += Math.abs(ll - v[i][second]); ul = v[i][second]; ll = v[i][second]; } } return ans;}// driver code let v = [[ 1, 3 ], [ 2, 5 ], [ 6, 8 ], [ 1, 2 ], [ 2, 3 ]]; let n = 5; document.write(solve(v, n)); </script> |
Output:
7
Time Complexity: O(N)
Auxiliary Space: O(1)
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